1
Measurements and SI units
Objectives
At the end of chapter 1, you must be able to:
Discuss quantities and their SI units
Discuss prefixes for SI units
Describe standard notation
Convert one unit to another
Take scale reading of various measuring
instruments
1.1 Units of measurements
In Book 1 we covered the SI units of basic quantities and derived quantities. All measurements
are represented using the international system of units, Standard International Unit (SI Unit).
Table 1.1 shows the summary of SI units of measurements of the quantities:
Table 1.1 Quantities and their SI units
Quantity
Name of SI base
unit
Symbol for SI base
unit
Length
Metre
m
Mass
Kilogram
Kg
Area
Square metre or
metre squared
m
2
Time
Second
s
Volume
Cubic metre or metre
cubed
m
3
Current
Ampere
A
Voltage
Volt
V
Resistance
Ohm
Ω
Temperature
Kelvin
K
Angle
Degree
o
Pressure
Pascal
Pa
Density
Kilogram per cubic
meter
Kg/m
3
CHAPTER 1
2
1.2 Prefixes for use with SI units
Some numbers are bigger or smaller. These numbers are shortened by an extra symbol called a
prefix. Table 1.2 below shows the prefixes for SI units.
Table 1.2 prefixes for SI units
Prefix
Symbol
Meaning
Example
Tera
T
1000 000 000 000 (1x10
12
)
Terabyte (TB)
Giga
G
1000 000 000 (1x10
9
)
Gigabyte ((GB)
Mega
M
1000 000 (1x10
6
)
Megawatt (MW)
kilo
k
1000 (1x10
3
)
kilometre (km)
deci
d
1/10 (1x10
-1
)
decimetre (dm)
centi
c
1/100 (1x10
-2
)
centimetre (cm)
milli
m
1/1000 (1x10
-3
)
millimetre (mm)
micro
μ
1/1000 000 (1x10
-6
)
microsecond (ms)
nano
n
1/1000 000 000 (1x10
-9
)
nanosecond (ns)
Speed
Metres per second
m/s
Velocity
Metres per second
m/s
Acceleration
Metres per second
per second
m/s/s or m/s
2
Force
Newton
N
Work
Newton metre or
Joule
Nm or J
power
Watt
J/s or W
Period
Second
s
Frequency
Hertz
Hz
Exercise 1.1
In your groups, answer the following questions:
1. State the SI unit for:
a. length b. time c. mass d. power
2. List three
a. derived quantities b. basic quantities c. basic SI units d. derived SI units
3
1.3 Standard notation
In Table 1.2, the numbers have been expressed in powers of 10.
For example: 1 Megawatt = 1000 000 watt = 1x10
6
. Numbers written using powers of 10 are said
to be in standard notation or scientific notation or standard form. Table 1.3 shows examples
of some standard notation.
Table 1.3 standard notation
Standard notation
1x10
12
1x10
9
1x10
6
1x10
-1
1x10
-2
1x10
-3
1.4 Scale reading of measuring instruments
Measurement of length
The SI base unit of length is the metre (symbol m). Other units of lengths are as shown below:
1 kilometre (km) = 1000 m = 1x10
3
m
1 centimetre (cm) = 1 m = 1x10
-2
m
100
Exercise 1.2
In your groups,
1. Write the following in standard notation:
a. 1000 000 000 000 000 w b. 10 000 000 000 g
c. 1/1000 000 000 000 s d. 0.000 0001 m
2. Convert the following to watts (w)
a. 20 Kw b. 4 Mw c. 1000μw
4
1 millimetre (mm) = 1 m = 1x10
-3
m
1000
1 micrometre (μm) = 1 m = 1x10
-6
m
1000 000
1 nanometre (nm) = 1 m = 1x10
-9
m
1000 000 000
Instruments for measuring length
Rule (ruler)
A metre rule is used to measure the lengths of distances between 1mm and 1metre.
The scale on the rule is found by checking the number of divisions between two values. On this
ruler there are 10 divisions between 0 cm and 1 cm. The scale is found as:
Scale = 1cm
10
Therefore, the scale is 0.1 cm.
Or we can use 10 divisions from 0 mm to 10 mm,
Scale = 10 mm
10
Therefore, the scale is 1 mm.
Vernier calipers
Vernier calipers are used when smaller and accurate measurements are required. Vernier
calipers consist of two parts:
a. The main scale which is fixed. It is usually numbered in cm.
b. The Vernier scale, the part that slides along the main scale. It has 10 divisions, each
0.9 (9/10) mm. The scale gives readings to 0.1 mm or 0.01 cm.
Figure 1.1 a rule (ruler)
mm 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160
5
How to use the Vernier calipers
The object whose length is required is placed between the jaws. Close the jaws onto the object to
be measured. Read the main scale, e.g. 1.4 cm. Identify the mark on the Vernier scale which
coincides exactly with a mark on the main scale, e.g. 0.3 mm or 0.03 cm. Take this reading to
give a second decimal place. The reading will be found as 1.4 cm + 0.03 cm = 1.43 cm.
Micrometer screw gauge
The micrometer screw gauge, shown in Figure 1.3, is used to measure accurately the dimensions
of all small objects.
How to use micrometer screw gauge
Rotate the thimble until the wire is firmly held between the anvil and the spindle.
To take a reading, first look at the main scale. This has a linear scale reading on it. The long lines
are every millimetre and the shorter ones denote half a millimetre in between. The scale on the
Figure 1.3 micrometer screw gauge
6
linear scale is 0.5 mm or 0.05 cm. The rotating scale is 0.01 mm or 0.001 cm. Then look at the
rotating scale. Add the 2 numbers, on the scale on the right.
From Figure 1.3:
Sleeve reads = 8 mm or 0.8 cm
Thimble reads = 0.12 mm or 0.012 cm
Total reading = 8.12 mm or 0.812 cm
Measurement of mass
The Mass of a substance is the quantity of matter contained in the substance. The SI base unit
for mass is the kilogram (kg).
Other units of mass are as shown below:
1 tonne (t) = 1000kg = 1x10
3
kg
1 gram (g) = 1 kg = 1x10
-3
kg
1000
1 milligram (mg) = 1 kg = 1x10
-6
kg
1000 000
Mass is measured by the instruments shown below:
Triple beam balance
Figure 1.4 instruments for measuring mass
Measurement of time
The SI base unit of time is the second (s).
Other units of time are as follows:
1 millisecond (ms) = 1 s =1x10
-3
s
1000
Top pan balance
7
1 microsecond (μm) = 1 s =1x10
-6
s
1000 000
1 nanosecond (ns) = 1 s =1x10
-9
s
1000 000 000
Time is measured by clocks and watches.
The time intervals are found by using a stop watch.
Experiment 1.1
AIM: To measure time intervals using a stop watch.
MATERIALS: Stop watch, meter rule, 50g mass, clamp stand, clamp and a string.
PROCEDURE:
1. Set up the apparatus as shown in Figure 1.5 below.
Figure 1.5
2. Pull the mass to one side at an angle of about 10
0
and leave it to vibrate freely.
3. Start the stop watch after one or two oscillations.
4. Read and record the time taken to make 10 complete oscillations.
5. Repeat the experiment using lengths 30 cm, 20 cm and 10 cm.
Compare your results with your friends’.
8
Measurement of volume
Volume is the quantity of space an object occupies. The SI base unit for Volume is metre cubed
(m
3
)
Volume can also be measured using the centimetre cubed (cm
3
)
1cm
3
= 1 m
3
= 10
-6
m
3
1000 000
1000 000cm
3
= 1m
3
Volumes of regular solids
For a regular block, volume = length x width x height
For a cylinder, volume = base area x height = πr
2
h
For a sphere = 4/3πr
3
Exercise 1.3
In your groups, answer the following questions:
1. State the situation where you need the following instruments:
a. Vernier calipers b. micrometer screw gauge
2. Explain how the following instruments work
a. Vernier calipers b. micrometer screw gauge
3. Liz wants to find the thickness of her smallest finger. Explain how best she can do it.
4. Kelson wants to find the speed of an athlete. Complete the table below for the instruments
he should use and the quantities he will measure.
Instrument
Quantity
9
Volumes of an irregular solid
Volume = the volume of a displaced liquid in a measuring cylinder
Volume of an irregular solid = volume B - volume A
= 125cm
3
– 75cm
3
Volume of an irregular solid = 50cm
3
A B
Figure1.6 measuring a cylinder used to measure the volume of an
irregular solid
Experiment 1.2
AIM: To measure volume using a measuring cylinder
MATERIALS: water, measuring cylinder, thin string and 3 stones of different sizes
PROCEDURE:
1. Pour water in the measuring cylinder about half-full.
2. Read and record the volume of water as VA.
3. Insert a stone tied to a thin string in the water.
4. Read and record the new volume of water as VB.
5. Calculate the volume of the stone by using the formula, V = VB – VA.
6. Repeat the experiment with the other two stones.
10
Volume of a liquid
The volume of a liquid is measured in litres.
Other units of volume of a liquid are as follows:
1 litre = 1000 cm
3
1dm
3
= 1000 cm
3
1l = 1dm
3
1l = 1000 ml
1cm
3
= 1 ml
Volume of a liquid is found by pouring the liquid in a measuring cylinder.
Figure 1.7 measuring cylinder
11
Other instruments that can be used to measure volume of a liquid are:
Pipette: for getting fixed pre-determined volumes.
Burette: delivers any volume up to its total capacity.
Summary
The SI units of the quantities are shown in the table below:
Quantity
Name of SI base
unit
Symbol for SI base
unit
Length
Metre
m
Mass
Kilogram
Kg
Area
Square metre or
m
2
Pipette
Figure 1.8 volumetric glassware
Burette
Exercise 1.4
In your groups, answer the following questions:
1. Write the following in litres:
a. 10 cm
3
b. 10 m
3
c. 100 ml
2. A cuboid has measurements 12 cm by 10 cm by 10 cm. Calculate its
volume in:
a. cm
3
b. m
3
c. l
3. Describe how you can find the volume of an irregular solid.
12
Prefixes are used in larger and smaller quantities as shown in the table below.
Prefix
Symbol
Meaning
Example
Tera
T
1000 000 000 000 (1x10
12
)
Terabyte (TB)
Giga
G
1000 000 000 (1x10
9
)
Gigabyte ((GB)
Mega
M
1000 000 (1x10
6
)
Megawatt (MW)
kilo
k
1000 (1x10
3
)
kilometre (km)
deci
d
1/10 (1x10
-1
)
decimetre (dm)
centi
c
1/100 (1x10
-2
)
centimetre (cm)
milli
m
1/1000 (1x10
-3
)
millimetre (mm)
micro
μ
1/1000 000 (1x10
-6
)
microsecond (μs)
nano
n
1/1000 000 000 (1x10
-9
)
nanosecond (ns)
The numbers that are written using powers of 10 are in scientific notation or standard form.
Length is measured in metres by a rule (ruler). Length of several metres can be measured by
using a tape measure. Length or thickness of small objects can be measured by using a
micrometer and Vernier calipers
Mass is measured in kilograms by using a top pan balance and triple beam balance.
metre squared
Time
Second
s
Volume
Cubic metre or metre
cubed
m
3
Current
Ampere
A
Voltage
Volt
V
Resistance
Ohm
Ω
Temperature
Kelvin
K
Angle
Degree
o
Pressure
Pascal
Pa
Density
Kilogram per cubic
meter
Kg/m
3
Speed
Metres per second
m/s
Velocity
Metres per second
m/s
Acceleration
Metres per second
per second
m/s/s or m/s
2
Force
Newton
N
Work
Newton metre or
Joule
Nm or J
power
Watt
J/s or W
Period
Second
s
Frequency
Hertz
Hz
13
Volume of a regular solid is measured in cubic metres by using the formulae, v = l x w x h.
Volume of an irregular solid is measured by using the volume of the displaced water in a
measuring cylinder.
Volume of a liquid is measured by pouring water in the measuring cylinder.
Student assessment
1. Write down the value of
a. 1040 cm in m
b. 700 g in kg
c. 30000 μs in s
2. The units of measurements are given as follows:
m g s mm ns cg cm
3
ms μm ml kg km
a. which of the above are
i. Units of mass?
ii. Units of volume?
iii. Units of time?
iv. Units of length?
v. Units of density?
b. Which of the above are measured by
i. A metre rule?
ii. A scale?
iii. A stopwatch?
iv. A measuring cylinder?
v. A vernier caliper?
vi. A micrometer?
3. Express the following in standard form:
a. 410 000 000 000 mm
b. 302 000 l
c. 23 007 g
d. 0.0000782 m
e. 0.0000009 cm
3
f. 0.00000089 s
4. A wooden block measures 10 m x 10 m x 8 m. Calculate its volume.
14
5. A cylindrical can has a height of 10 m and area 40 cm
2
. What is the radius of the can?
6. The volume of the water in the measuring cylinder is 40 cm
3
. A stone is lowered in the
cylinder and its volume rose to 100 cm
3
. Calculate the volume of a stone in
a. cm
3
b. m
3
7. Write down the Vernier reading in Figure 1.9:
8. What is the reading on the micrometer screw gauge?
9. Explain how you can find the volume of
a. A cube
b. A stone
10. Name the instrument you would use for the following measurements:
a. Thickness of a coin
b. Length of a building
c. Time intervals
d. Time of the day
Figure 1.9
Figure 1.10
15
e. Your height
f. Length of your exercise book
g.
11. Write down the volume of a solid in Figure 1.11.
12.
Figure 1.11
16
Scientific investigations
Objectives
At the end of chapter 2, you must be able to:
Design a scientific investigation
Carry out a scientific investigation
Analyse data from a scientific investigation
Communicate results from experiments
Evaluate a scientific investigation
2.1 Designing a scientific investigation
Designing a scientific investigation involves the following stages:
Identifying a problem
Hypothesising
Deciding the type of data to collect
Identifying variables
Identifying a problem
In an investigation, start with the problem that you want to investigate. This is like a question
that you need answers for. For example: what is the effect of voltage on current in the circuit?
Hypothesising
This is the stage where you make a prediction. The prediction is called hypothesis.
For example: current in the circuit increases when voltage increases. The hypothesis may not
be right. Therefore, this prediction is tested during an investigation.
Chapter 2
17
Deciding the type of data to collect
The data to be collected during an investigation must be decided before carrying out the
investigation. For example,
Decide the range of voltage readings to be collected
Decide the range of current values to be collected
Identifying variables
In this case we identify what is going to be observed or measured. These are called variables.
Variables can be defined as factors that would affect the results of the investigation.
Variables can be anything that can change. Variables are mainly taken from the hypothesis.
Variables in this investigation are number of cells, voltage and current.
Independent variable: The variable that you are changing in an investigation or experiment.
This variable affects what happens in an investigation. In this investigation
the independent variable is number of cells. Changing the number of cells
will change the amount of voltage and current in the investigation.
Dependent variable: This is what you will be measuring. For example: voltage and current.
Control variables
During an investigation, some variables do not have to be measured. These variables need to be
controlled. For example: temperature and the value of resistor.
Exercise 2.1
In your groups, discuss how you can design a scientific investigation on
how the frequency of a pendulum depends on the length of the string.
18
2.2 Carrying out a scientific investigation
RESULT
The voltmeter and ammeter readings increase when the number of cells increases.
EXPLANATION
The voltage increases because the force pushing electrons in the circuit increases. This causes an
increase in the amount of current in the circuit.
CONCLUSION
Therefore, an increase in voltage causes an increase in current.
Experiment 2.1
AIM: To investigate the effect of voltage on current
MATERIALS: Connecting wires, ammeter, voltmeter, 4 cells, switch and resistor
PROCEDURE:
1. Set up the experiment as shown in Figure 2.1.
2. Close the switch and take the voltmeter and ammeter readings.
3. Repeat the experiment with 2, 3 and 4 cells.
4. For each number of cells, take the voltmeter and ammeter readings.
5. Record the results in table 2.1 below:
Number of cells
Voltmeter reading (V)
Ammeter reading (A)
1
2
3
4
Table 2.1
6. Discuss your results with other students in your class.
Figure 2.1
R
19
Controlling variables
In experiment 2.1, temperature and the resistor are the variables that are controlled.
To get a fair result, you should change one variable (e.g. number of cells) at a time and check
how it affects other variables (e.g. voltage and current).
Collecting scientific data
Make sure you have written the observations properly. State the unit in which each
measurement is made, for example 0.1 A for current.
You can use a table. Make sure you enter the observations in the table and indicate the units at
the top of each column only. In a table draw rows and columns. Rows are horizontal gaps while
columns are vertical gaps. Place independent variables (this is what you are changing in the
experiment) in the first column. Place the dependent variables (this is what you will be
measuring) in the next column (s).
Table 2.2 below shows the results for an experiment like experiment 2.1.
No measurement is exact. There is always some uncertainty about it. For example, if the values
are like in Table 2.3, it is very important to give your calculations to an approximate number of
significant figures.
Voltage (V)
Current (A)
Resistance (Ω) = Voltage (V)
Current (A)
1.0
0.15
6.666666667
3.0
0.48
6.25
5.0
0.81
6.17283906
Table 2.3
In this case the measurements of voltage and current are given to 2 significant figures. Therefore,
the calculations for resistance should also be rounded off to 2 significant figures.
Independent variable Dependent variables
Number of cells
Voltage (V)
Ammeter (A)
1
1.5
0.1
2
3
0.2
3
4.5
0.3
4
6
0.4
Table 2.2
20
Voltage (V)
Current (A)
Resistance (Ω) = Voltage (V)
Current (A)
1.0
0.15
6.7
3.0
0.48
6.3
5.0
0.81
6.2
Table 2.4
2.3 Analysing data from a scientific investigation
Errors and accuracy during experiments
The results of an experiment can be slightly inaccurate for two main reasons:
1. You can make personal errors in the observations.
2. The apparatus itself can be capable of only limited accuracy.
The errors can be grouped into:
1. Personal errors
The common personal error is due to parallax. Parallax error is the apparent change in the
position of an object due to a change in the position of your eyes and every time you measure a
length or read a pointer moving over a scale it is likely to arise.
Figure 2.2 shows how the position to be read varies, with respect to the scale, as the eye is
moving from P1 to C to P2.
Positions P1 and P2 are wrong while position C is only correct one.
Figure 2.2 Taking a reading from a ruler
C
P1
P2
21
When reading a liquid level, you take the reading from the meniscus. The meniscus should
always be viewed horizontally to avoid parallax as shown in Figure 2.3.
2. Errors of the instrument
If you are using an apparatus which is not sufficiently sensitive, it is difficult to produce a good
experimental result. For example, it is difficult to time a race using a watch with no second hand.
So you will be unable to produce a good experimental result with apparatus which is not
sufficiently sensitive. Another example can be a measuring cylinder. Consider the determination
of the volume of a solid by displacement of water in a measuring cylinder. In a 100 cm
3
cylinder
every cm
3
is marked, but the graduations on a 500 cm
3
vessel are only every 5 cm
3
. In this case,
the smallest cylinder into which the solid will go should be used because a more accurate reading
is possible.
Zero error: The error which occurs when the measuring instrument does not indicate zero when
it should.
3. Reading and recording
Estimation of reading to one-tenth of the smallest scale division is often necessary and should be
practiced. Always imagine the division divided up into ten equal parts and estimate which tenth
coincides with the mark to be read.
Figure 2.3 Taking the reading from the meniscus
22
The reading on an ammeter is 0.155 A.
Minimisation of errors
You can produce accurate results from an experiment by reducing the errors.
Errors can be reduced by:
Taking an average of several readings. Therefore, you repeat the experiment.
Avoiding parallax error. Therefore, if you are reading from a scale, make sure you look at
right angles to it so that you read a correct number.
Avoiding the zero error. Therefore, make sure that the instrument is pointing at zero before it
is used in the experiment.
Figure 24 reading scale
23
Plotting the graph
The results can be plotted on the graph.
The following are the hints on drawing graphs:
Choose your scale so that the graph fills the paper, and label the axes.
Mark each point by a dot or with a cross.
Join only those points that are on the same lines in the straight line graph.
Use a ruler to draw an obviously straight line graph, putting the line in such a way that the
points are evenly distributed about it.
Write the title of the graph.
Exercise 2.2
In your groups, read the instruments shown below correctly:
Figure 2.5
1.
3
2
24
During an experiment, not all the points can be on a straight line. So you should draw a straight
line on a graph that goes through as many points as possible. This is called a line of best fit.
If a point is lying outside the range of the straight line, it is treated as an error and do not include
it when drawing the straight line. For example: point P.
Title: Voltage against current
Figure 2.6 graph of voltage against current
Title: Voltage against current
Figure 2.7 line of best fit
25
If points are scattered, you can draw the graph of the best fit which is an average of all the points
as shown in Figure 2.8.
Conclusion from the graph
Give a clear conclusion in simple and straight forward sentences. Be honest when writing the
account and conclusion.
From the graph in Figure 2.6, you can draw the conclusion depending on the shape of the graph.
We can observe that the current in the circuit increases with an increase in voltage or we can say
that the current in the circuit decreases with a decrease in voltage. When voltage is doubled
current is also doubled. This is only true when temperature and other physical factors are kept
constant. Therefore, voltage and current are in direct proportion.
2.4 Communicating results from experiments
You can communicate your results from an experiment by including the following points:
Organizing results from the experiment
Making oral and poster presentation of the findings
Sketching and labeling experimental set up
Writing laboratory report.
Figure 2.8 graph of best fit
26
Laboratory report
Aim
The aim of this investigation was to determine the relationship between the length of nichrome wire and its
resistance.
I knew that I can find the resistance of the nichrome wire by finding the voltage and current across the
nichrome wire. Then I can work out the resistance by using a formula shown below:
Resistance (R) = Voltage (V)
Current (A)
In this case I would connect a nichrome wire in a circuit then measure voltage across it and current in the
circuit. I would do this for different lengths of nichrome wire.
Hypothesis
Length being one of the factors that affect resistance of a wire, it means when the length of the wire is
changing its resistance will also change. My prediction was that the resistance of the wire will increase when
the length of the wire increases and vice versa.
Variables
The key variables in this experiment were:
Length of nichrome wire: measured by a metre rule in centimetres (cm).
Voltage: measured by a voltmeter in volts (V)
Current: measured by an ammeter in amperes (A).
The variables that I controlled were:
Temperature: this was controlled by connecting the nichrome wire in a beaker of cold water.
Temperature must be controlled because when current flows through a nichrome wire it produces heat
and this heat affects the resistance of the wire.
Diameter (thickness) of nichrome wire: this was controlled by using the wire of the same thickness. So
in my investigation I was using the same wire and simply changing its length. Thickness of the wire
must be controlled because it affects the resistance of the wire.
Materials
The materials that I used during this investigation were: 100 cm nichrome wire (0.28 mm diameter),
ammeter(0-3A), voltmeter(0-6V), metre rule, connecting wires, crocodile clips, cold water, beaker, cells (battery)
and a variable resistor.
27
Procedure
I set up the experiment as shown below:
I connected a 100 cm nichrome wire in the circuit and recorded the voltmeter and ammeter readings.
I repeated the experiment with 80 cm, 60 cm, 40 cm and 20 cm nichrome wires. For each length, I recorded the
voltmeter and ammeter readings.
I recorded the results in the table as shown below:
Length of the wire (cm)
Current (A)
Voltage (V)
Resistance(Ω) = voltage(V)
current(A)
100
1.0
4.2
4.1
80
1.3
4.2
3.2
60
1.8
4.2
2.4
40
2.5
4.2
1.7
20
4.7
4.2
0.9
Safety: When I was carrying out this experiment, I made sure that the power supply was switched off before I
removed the nichrome wire to change its length.
I used my results to calculate the resistance of each length of the wire.
28
Drawing the graph
I used the values in the table to plot a graph of resistance against length with on the horizontal axis because it
is the independent variable and resistance on the vertical axis because it is a dependent variable.
The points on my graph are a little bit scattered but I have used a line of best fit which is a straight line.
Conclusion
From my prediction I expected the graph of resistance against length to be a straight line, which showed that
the resistance of the nichrome wire increases as the length of the wire increases.
Therefore, I can conclude that resistance of the wire is directly proportional to its length. This agrees with my
original hypothesis that doubling the length of nichrome wire also doubles its resistance and vice versa.
Evaluation
The points on my graph are uneven but I am sure they would lie on a straight line.
There are reasons why my points may have been scattered. Some of the reasons are:
Personal error due to parallax
Scale on the instruments
To get more accurate results I would have done the following:
Repeated the experiment to get average results that are accurate
Avoid parallax error – make sure I look at a right angle to the scale of the instrument so that I read a
correct value
29
Avoid the zero error- make sure that the instrument is pointing at zero before it is used in the
experiment.
2.5 Evaluating a scientific investigations
An evaluation helps to decide how reliable your conclusions are. It also helps how the
experiment could be improved. In examinations, you may be asked to comment on how precise
or reliable the evidence is. You may also be asked how to improve that accuracy or reliability.
Reliability
You must comment about uncertainties in your measurements. This can be the reliability of the
readings, especially in relation to the scale of the measuring apparatus.
In an experiment, you will find some results which do not agree with the others. These results
look like mistakes and we call them anomalous results.
The reliable results must be the results that if the measurements are repeated, the same result
should be obtained. On a best fit line, the reliability of the results can be checked by checking the
closeness of the points to the line. When most of the points are very close to the best fit line, we
say the results are reliable.
Ways of reducing factors that may affect a scientific
investigation
After completing the experiment, you must suggest ways of improving it. This is important in
order to have more reliable conclusion.
1. Precision
To be precise means that the measurements were done as accurately as possible.
For example
If you are carrying out an experiment to measure very small quantities, e.g. voltage, you may use
a millivoltmeter instead of a voltmeter in order for the results to be more precise.
Exercise 2.3
In your groups, prepare a presentation of the results you obtained in
experiment 2.1.
30
If you are measuring time intervals, e.g. time taken for one oscillation of a pendulum to be
performed, it is very important to record the time for 10 oscillations. Divide the total time for 10
oscillations by 10. In this case, the errors by the human reaction time are minimized because they
are spread out over many oscillations.
2. Reliability
Reliability of the results can be improved by repeating the experiment and compare the results.
The results are close to each other, and then the results are reliable.
Summary
Designing of a scientific investigation involves identifying a problem, hypothesising, deciding
the type of data to collect and identifying variables.
When carrying out a scientific investigation some variables must be controlled and just change
one variable in order to have a fair test. Then collect and organize the scientific data e.g. in a
table.
In an experiment, the results can have some errors because of personal errors, instrumental errors
and recording errors.
These errors can be minimised by repeating the experiment, avoiding parallax errors and
avoiding zero error.
When drawing a graph, you must choose axes, choose scales, label the axes and draw the best
line.
Conclusion is made from the trend of the graph.
Evaluation is used to help you decide how reliable your conclusions are and how your
experiment could be improved or extended.
Exercise 2.4
In your groups, use the results in experiment 2.1 to:
1. Decide how far the results support the hypothesis.
2. Decide if your results were reliable.
3. Discuss shortcomings that may affect the results.
4. Discuss ways of reducing factors that may affect the results of the experiment.
31
Student assessment
1. Define the following terms:
a. Parallax error
b. Zero error
2. What instructions would you give to enable accurate measurements to be made from a
measuring cylinder?
3. List down three hints on drawing graphs.
4. During a physics practical, a student took the reading of a voltmeter. The reading was 0.52V.
The required reading was 0.51V. What was the experimental error?
5. Frequency of an oscillating cantilever (ruler) is affected by its length. Table 2.5 below shows
the results obtained during an investigation.
Table 2.5
Length of a cantilever (cm)
10
20
30
40
50
Frequency (Hz)
1.8
1.6
1.4
1.2
1
a. Use the results in Table 2.5 to draw a graph of frequency (y axis) against length of a
cantilever (x axis).
b. How can you verify that the results during the investigation were correct?
c. What is the relationship between the length of the cantilever and the frequency?
6. Explain what is involved in:
a. hypothesising
b. controlling variables
c. conclusion
d. evaluation
7. What is the importance of controlling variables during a scientific investigation?
8. In a scientific investigation,
a. Discuss three classifications of errors.
b. Discuss ways of minimising the errors mentioned in 8(a).
9. Discuss what must be included in a written report of the investigation.
32
10. Write down the readings on the following instruments:
11. A form three class at Phunziro Secondary School was investigating the change in
temperature of hot water (100 cm
3
) as cold water was added.
a. Name the instrument that can be used to measure the temperature of water.
b. Apart from the instrument in 11 (a), what are the other required apparatus?
c. The results obtained during the investigation were recorded in the Table 2.6:
Table 2.6
Volume of cold water added
(cm
3
)
Temperature (
0
C)
0
100.0
20
78.0
40
68.0
60
60.5
80
54.0
100
39.0
Figure 2.9
a
b
c
33
d. Use the data in the table to plot a graph of temperature (y-axis) against volume (x axis).
Figure 2.10
e. What conclusion can you make from this investigation?
12. The results obtained to find the relationship between length of nichrome wire and resistance
are shown in Table 2.7.
Table 2.7
Length of the wire
(cm)
Current (A)
Voltage (V)
Resistance(Ω) =
voltage(V)
current(A)
100
0.2
1.2
80
0.4
1.2
60
0.6
1.2
40
0.8
1.2
20
1.0
1.2
a. Calculate the resistance of each length of the wire and complete the table.
b. Theory states that resistance of the wire is directly proportional to its length. State
whether your results support this theory. Justify your answer by reference to the results.
c. State three variables that were kept constant during this investigation.
34
d. Suggest one precaution you could take to ensure that the readings are as accurate as
possible.
13. Figure 2.11 shows points marked on a graph from the results obtained during an
investigation.
Figure 2.11
a. Draw a smooth curve by joining the points.
b. Which point shows an inaccurate result?
c. Give a reason to your answer in 13 (b).
14. Evelyn is investigating the effect of surface area exposed to the air on the rate of cooling of
hot water. She designed the apparatus as shown in Figure 2.12.
Volume of gas
(cm
3
)
35
a. Name two variables that are kept constant.
b. Name one variable which is changing.
c. Give one hypothesis of the investigation.
15. Figure 2.13 is a graph of voltage against current
a. Explain the use of plotting a line of best fit.
Voltage
(V)
Current (A) Figure 2.13
36
b. What conclusion can you draw from the graph?
c. State the value of
i. Current when voltage is 1.5 V
ii. Voltage when current is 0.7 A.
d. Suggest shortcomings that might affect the results of a scientific investigation.
e. Discuss ways of reducing factors that may affect a scientific investigation.
37
Kinetic theory of matter
Objectives
At the end of chapter 3, you must be able to:
Describe the kinetic theory of solids, liquids and
gases
Explain the cause of gas pressure
Explain the relationship between average
molecular speed and temperature
Explain the meaning of absolute temperature
3.1 Three states of matter
What is matter? Matter covers all the substances and materials from which the physical universe
is composed. Matter is anything which has mass and volume or occupies space. All the
substances and materials are categorised as solids, liquids and gases. Therefore, the three states
of matter are solid, liquid and gas.
Particle arrangement in the three states of matter
Solids: Solids have very strong intermolecular forces. Their particles are closely packed in a
regular pattern. Their particles vibrate within a fixed point when heated.
Liquids: Liquids have weak intermolecular forces compared to solids. Their particles not closely
packed and they slide over each other because there are spaces between them.
Gases: Gases have weakest intermolecular forces. Their particles are further apart and move
freely because there are larger spaces between them.
Chapter 3
38
3.2 The kinetic theory
Kinetic theory is a scientific explanation of the behaviour of these three states of matter. It is a
theory which accounts for the bulk properties of matter in terms of constituent properties.
The main points of the kinetic theory are:
All matter is composed of smaller particles (molecules, atoms or ions) which have
different sizes. These particles are invisible to the naked eye.
The particles are held together by intermolecular forces (IMFs). Intermolecular forces
are forces of attraction between particles of a state of matter.
Factors that affect the size of intermolecular force in the given state of matter are:
Distance between particles
An increase in distance d between particles decreases the strength of intermolecular forces and
vice versa.
The size of the particles
Increasing the size of the particles increases the strength of intermolecular forces and vice versa.
d d
Stronger intermolecular forces weaker intermolecular forces
Figure 3.2 effect of distance on the size of intermolecular forces
Figure 3.1 arrangement of particles in the three states of matter
39
Kinetic theory also states that the vibrations of the particles become greater as the
temperature rises.
From Experiment 3.1, compare your results with the following results:
3: The balloon contracts because particles lose kinetic energy and move closer to each other.
5: The balloon expands because particles gain kinetic energy and move further apart.
CONCLUSION
From the results in experiment 3.1, we can conclude that particles in a state of matter are always
in motion.
Weak intermolecular forces
Strong intermolecular forces
Figure 3.3 effect of size of the particles on intermolecular forces
Experiment 3.1
AIM: To investigate the kinetic molecular theory of matter
MATERIALS: Balloon, string, heat source and fridge (freezer)
PROCEDURE:
1. Inflate a balloon.
2. Put the balloon in a freezer for some time.
3. What happens to the size of the balloon? Explain in terms of kinetic
theory.
4. Remove the balloon and put it near the heat source or under the sun.
5. What happens to the size of the balloon? Explain in terms of kinetic
theory.
Discuss your results with your friends in class.
40
3.3 Properties of matter
Properties of Solids
Solids have the following properties:
They have a fixed shape because solids are crystalline and atoms in it are set in well –
defined patterns.
They have a fixed volume.
They have high density because their molecules are held closer to each other.
The volume occupied by particles is less compared to other states.
They are incompressible because there are no spaces between particles of solids.
Their particles vibrate about a fixed mean position.
Their particles’ vibration increases as temperature increases and their separation increases
slightly.
Properties of Liquids
Liquids have the following properties;
Their particles are closer to each other but relatively further apart when compared to
solids.
They take the shape of a container which holds them because their molecules slide over
each other.
They have a fixed volume.
They cannot be compressed because the spaces between molecules are very small.
Molecules in liquids vibrate more and they move at a very high speed throughout the
body of the liquid.
Properties of gases
The properties of gases are as follows:
They do not have a fixed shape because their molecules are far apart and there are a lot of
free spaces between them.
They take the shape of the container.
They do not have fixed volume because their molecules can easily escape; therefore, they
take the volume of the container which holds them.
They have low density because they occupy a greater space (greater volume).
They can easily be compressed because there are a lot of spaces between molecules.
41
Their molecules move randomly and at quite high speeds at normal temperature and
pressure.
3.4 Gas pressure
Pressure is defined as force exerted per given area. In gases, pressure is caused by the force
exerted by gas molecules per given area on the surface of the container.
From experiment 3.1, when the balloon was placed near the heat source, it expanded. The
pressure inside the balloon is caused by the gas particles striking the walls of the balloon. An
increase in temperature causes an increase in kinetic energy of the particles. The particles that
have more energy move faster and strike the inside surface of the balloon more frequently. This
causes an increase in pressure.
Gas pressure can also increase by increasing the number of molecules of the gas.
When using a pump to inflate a balloon, the number of gas molecules increases. This increase in
gas molecules makes the molecules to strike the inside surface of the balloon most frequently in
all directions. This causes an increase in pressure. Hence the balloon expands.
Exercise 3.1
In your groups, answer the following questions:
1. Draw diagrams to show the arrangement of particles in:
a. Molten sodium chloride
b. Gaseous water
c. An ice block
2. Explain, using kinetic theory, the following:
a. Particles in gaseous lead diffuse faster than particles in molten lead.
b. Particles in solids vibrate within a fixed point.
c. The balloon containing air expands when heated.
42
3.5 Molecular motion and temperature.
Figure 3.6 shows a sketch of the graph from experiment 3.2
The kinetic theory of matter can be used to explain how a substance changes from one state to
another.
When a solid is heated, its temperature increases (between A and B). Its particles gain kinetic
energy and vibrate more, moving further away from each other. The heat energy supplied
between B and C weakens the intermolecular forces. This heat energy is called latent heat of
fusion. This breaks the regular pattern. The particles now move around each other. Solid forms a
Experiment 3.2
AIM: To determine the melting point of ice and boiling point of water.
MATERIALS: Bunsen burner, matches, ice blocks, tripod stand with wire gauze,
thermometer, beaker, stop watch and graph paper.
PROCEDURE:
1. Put ice blocks in a beaker. Measure the temperature of ice blocks.
2. Light a Bunsen burner and heat the ice blocks.
3. Record the temperature changes every minute until the ice blocks melt and the
liquid boils. Record your results in the table.
Time(min)
1
2
3
4
5
6
7
8
9
10
Temperature(
0
C)
Table 3.1
4. Plot a graph of temperature (
0
C) against time (min).
Discuss your observations with your friends in class.
Write down the melting point of ice and the boiling point of water.
Figure 3.6 the heating graph of matter
43
liquid and the process is called melting. This takes place at a constant temperature called
melting point. The temperature remains constant while melting because all the heat energy
supplied is used to break down the internal bonds.
When a liquid is heated its temperature starts increasing again (between C and D). Its particles
move faster by gaining kinetic energy. The particles that are on the surface have enough energy
to overcome the forces between themselves. These particles of the liquid escape to form a gas.
The process is called evaporation. Further heating makes the particles to escape from the liquid
so quickly. The liquid starts boiling (between D and E). The temperature is called boiling point.
This temperature is also constant because the substance gained latent heat called latent heat of
vapourisation which is used to break down the internal bonds.
Table 3.2 shows the boiling and melting points of some substances
Substance
Melting point (
0
C)
Boiling point (
0
C)
Water
0
100
Aluminium
661
2467
Sulphur
113
445
Ethanol
-117
79
Magnesium oxide
2827
3627
Mercury
-30
357
Methane
-182
-164
Oxygen
-218
-183
Sodium hydroxide
801
1413
When a gas is cooled, its temperature decreases. The average kinetic energy of the particles
decreases and they move closer to each other. The intermolecular forces increase and this causes
the change of gas to liquid. The process is called condensation.
When a gas is cooled, its temperature decreases. The average kinetic energy of the particles
decreases and they move closer to each other. The intermolecular forces increase and this causes
the change of liquid to solid. The process is called freezing.
NOTE: During condensation and freezing, heat energy is given out.
44
Figure 3.7 is a summary of changes of state of matter.
Motion of gas molecules
Gas molecules move at high speed at random in a container. The movement is known as
Brownian motion.
When the temperature of the gas increases, the speed of molecules also increases because
particles gain kinetic energy. This is shown by the collisions or bombardment of invisible
molecules on the visible particles.
Diffusion
Diffusion is the movement of molecules (fluid molecules) from a region of high concentration to
a region of low concentration. This is the process by which different substances mix as a result of
the random motions of their particles.
Figure 3.7 changes of states of matter
Melting
Boiling
Condensation
Freezing
Figure 3.8 Brownian motion of gas molecules.
45
Diffusion stops when there is even distribution of the fluid.
If you open a bottle of perfume in one corner of a room the scent can be detected throughout the
whole room because the scent will move from a region of high concentration (where the bottle
is) to a region of low concentration (where there is no perfume).
Demonstrating diffusion
Diffusion of bromine gas and air can be demonstrated as shown in Figure 3.9.
When the glass plate is removed so that the two open ends of the jars are in contact, bromine gas
diffuses rapidly into the air.
Diffusion of bromine gas into the air is noticed by the paler brown colour of bromine in air.
This takes place until there is uniform paler brown colour in both jars. The air molecules again
diffuse into the bromine gas.
NOTE
Diffusion can take place at a high rate if it is carried out at high temperature because high
temperature increases the kinetic energy of the particles.
Lighter molecules diffuse faster than heavy molecules.
Diffusion also occurs in liquids but it takes much longer days because molecules in liquids are
not very fast as explained in kinetic theory of liquids
Figure 3.9: diffusion of bromine gas and air
46
It also occurs in solids. But diffusion is not noticeable in solids because it takes many years for a
very small layer of the substance to diffuse. This is so because molecules in solids are held close
together by strong forces.
Applications of diffusion
Diffusion has the following applications in the body:
Oxygen diffuses from alveoli (air sacs) into the blood capillaries in the lungs.
Carbon dioxide diffuses from the blood capillaries to the alveoli in the lungs.
Digested food diffuses from the small intestines into the blood capillaries of the villi.
3.6 Absolute temperature
Absolute temperature is the minimum temperature that any substance can reach when it is
cooled. If you put water in the freezer, the temperature of the water decreases and goes beyond
0
0
C. This substance’s temperature will stop decreasing when it reaches -273
0
C. This temperature
is called absolute zero. The Lord Kelvin proposed his temperature scale in 1854, called Kelvin
scale, which has 0 K at absolute zero.
When the temperature of water decreases, the kinetic energy of the particles also decreases. This
decreases the volume of water. At absolute zero the particles do not have any motion.
Therefore, we can also define absolute temperature as the temperature at which molecules have
the minimum possible kinetic energy.
Exercise 3.2
In your groups, answer the following questions:
1. Explain the cause of gas pressure.
2. With the aid of a diagram, describe the diffusion of copper sulphate in water.
3. Discuss the difference in diffusion in the three states of matter.
4. Explain why a solid melts when heated.
5. Draw and label the graph you would expect to produce if water vapour at 100
0
C
was cooled to a temperature of 0
0
C.
47
Figure 3.10 is a graph showing the relationship between the volume of a gas and temperature.
Summary
The three states of matter are solid, liquid and gas.
Kinetic theory explains why different states of matter behave differently.
Some properties of the states of matter are shown in Table 3.3:
Table 3.3
Solid
Liquid
Gas
Fixed shape
No fixed shape
No fixed shape
Fixed volume
Fixed volume
No fixed volume
Incompressible
Fairly high density
Low density
Incompressible
Easily compressed
When a substance is heated, its particles gain kinetic energy and move further apart to form a
new state of matter. Temperature remains constant when a substance changes its state because
the heat energy used is as latent heat.
Absolute temperature is the minimum temperature that a substance can reach when it is cooled or
the temperature at which the substance’s motion is minimum.
Diffusion is the movement of particles from a region of high concentration to a region of low
concentration.
Diffusion is fastest in gases because the spaces between particles are greater and the particles are
lighter than in other states of matter.
Volume (cm
3
)
-273
0
C 0
0
C Temperature
0K 273K
Figure 3.10 Volume of a gas against temperature
48
Student Assessment
1. Define the following scientific terms:
a. Intermolecular forces
b. Random motion
c. Diffusion.
2. State the difference between solids and liquids in terms of intermolecular forces.
3. Draw structures of a solid, liquid and gas to show the arrangement of their molecules.
4. Describe the differences between solids, liquids and gases in terms of the motion of the
molecules.
5. Describe an experiment which demonstrates Brownian motion of the smoke particles in
the air.
6. State two uses of diffusion.
7. State two differences between the process of evaporation and boiling.
8. Figure 3.11 shows a graph of temperature against time for a pure solid substance which
is heated until it changes into gas. Use it to answer the questions that follow.
a. Which parts of the graph corresponds to the substance existing in two states for each
part?
b. Explain how the heat supplied to the substance during B and D is used.
c. What happens to the substance at
i. B?
ii. D?
d. Which part of the graph shows a substance that has
i. Weakest IMF
Figure 3.11
49
ii. Strongest IMF
iii. Molecules with the greatest kinetic energy
9. Brownian motion of smoke particles in air is an example of diffusion.
a. Define ‘diffusion’.
b. State the effect in the rate of diffusion if the temperature increases.
10. Figure 3.12 is a diagram showing gases (hydrogen chloride and ammonia) from the solutions
diffusing along the tube and a white cloud forming where they meet at point A.
Figure 3.12
a. Which of the two gases is lighter?
b. Give a reason for the answer to 10 (a).
c. Which of the two gases is heavier?
d. Give a reason for the answer to 10 (c).
e. When the experiment is done on a sunny day, it takes a shorter time for the white cloud to
form. Give a reason.
11. a. Define the term Absolute temperature.
b. What happens to the molecular motion at absolute temperature?
12. Explain in terms of molecules
a. the process of evaporation.
b. the process of freezing.
c. how gas particles cause pressure on the walls of the container.
d. why air pressure inside a car tyre increases when the car is being driven.
13. With the aid of a diagram, describe the diffusion of nickel (II) sulphate in water.
50
14. Explain why diffusion in gases is faster than in liquid and solid.
15. When a candle wax is heated it melts. Explain why this happens.
16. When a person opens a bottle of ammonia gas, people in all parts of the room soon notice the
smell. Use kinetic theory to explain how this happens.
17. Table 3.4 shows the boiling points and melting points of some substances.
Substance
Melting point (
0
C)
Boiling point (
0
C)
Sulphur
113
445
Ethanol
-117
79
Mercury
-30
357
Methane
-182
-164
Sodium hydroxide
801
1413
Table 3.4
Which substance is
a. A gas at 25
0
C?
b. A liquid at 25
0
C?
c. A solid at 25
0
C?
18. Describe an experiment that you would carry out to demonstrate that lighter particles travel
faster than heavy particles.
51
Thermometry
Objectives
At the end of chapter 4, you must be able to:
Differentiate types of temperature scales
Describe how various thermometers function
4.1 Types of temperature scales
Measurement of temperature
Temperature is measured using instruments called thermometers. Temperature is measured
either in Kelvin (K) or Degrees Celsius (
o
C). Therefore, the two types of temperature scales are
Celsius scale and Kelvin scale.
Celsius temperature scale
A Celsius scale has a lower fixed point of 0
o
C and the upper fixed point of 100
o
C. Extensions
can be made above 100
o
C or below 0
o
C. The absolute zero temperature on the Celsius scale
is -273
0
C.
Kelvin temperature scale
Temperature is the measure of the hotness and coldness of a substance or an object. Temperature
is measured using instruments called thermometers.
Kelvin temperature scale is the scale found by Lord Kelvin in 1854. This is the scale which is
used by scientists in scientific work.
The lower fixed point of the Kelvin scale is at 273.15 K (approximately 273 K).
The upper fixed point of a Kelvin scale is 373.15 K (approximately 373 K). The absolute zero
temperature on the Kelvin scale is 0K.
Chapter 4
52
Relating Celsius scale and Kelvin scale
The changes in temperature on both the Celsius and the Kelvin scale are the same.
The two scales can be related as follows:
Lower fixed point on Celsius scale = lower fixed point on Kelvin scale
0
o
C = 273 K
Upper fixed point on Celsius scale = upper fixed point on the Kelvin scale
100
o
C = 373 K
Converting Celsius to Kelvin
If you want to convert degrees Celsius to Kelvin, you must add 273 to the temperature in degrees
Celsius.
Kelvin = degrees Celsius + 273
K =
o
C + 273
Worked examples
Convert the following degrees Celsius to Kelvin:
1. 100
0
C
2. -10
0
C
3. -273
0
C
Solutions
1. K =
0
C + 273
K = 100 + 273
K = 373 K
2. K =
0
C + 273
K = -10 + 273
K = 263 K
3. K =
0
C + 273
K = -273 + 273
K = 0 K
53
Converting Kelvin to degrees Celsius
If you want to convert Kelvin to degrees Celsius you must subtract 273 from the temperature in
Kelvin.
o
C = K – 273
Worked examples
Convert the following to degrees Celsius:
1. 350 K
2. 310 K
3. 0k
Solutions
1.
0
C = K - 273
0
C = 350 - 273
0
C = 77
0
C
2.
0
C = K - 273
0
C = 310 - 273
0
C = 37
0
C
3.
0
C = K - 273
0
C = 0 - 273
0
C = 273
0
C
Table 4.1shows a comparison of the Celsius and Kelvin scale.
Celsius (
o
c)
Kelvin (k)
Special occasion
100
373
Boiling point of water
78
351
Boiling point of ethanol
37
310
Normal body temperature
0
273
Melting point or freezing point of
water
-273
0
Absolute zero
54
4.2 Types of thermometers
Temperature is measured by an instrument called thermometer. There are different types of
thermometers depending on the physical property that varies with temperature.
Liquid-in-glass thermometer
Liquid-in-glass thermometer uses expansion and contraction of a liquid.
How a liquid-in-glass thermometer works
In the thermometer shown in Figure 4.1 above, mercury and ethanol liquids are used. The liquid
is in a thin capillary tube. When temperature increases, the walled bulb enables heat to pass
through quickly, and the liquid is heated. The liquid expands to attain the temperature of the
Exercise 4.1
In your groups, answer the following questions:
1. Convert the following to Kelvin:
a. -100
0
C b. 200
0
C c. 45
0
C
2. Convert the following to degrees Celsius:
a. 5K b. 450 K c. 15K
3. Explain why scientists prefer to use Kelvin scale over Celsius scale.
Figure 4.1: a liquid-in-glass thermometer
55
surrounding. When temperature decreases the liquid contracts and gives the temperature of the
surrounding.
A bulb of a thermometer is thin in order to enable heat energy to pass through quickly.
The capillary tube is very narrow so that a small change in temperature causes a reasonable
movement of the liquid. The glass stem is thick in order to prevent the glass from breaking. The
glass also acts as a measuring glass. Examples of liquid-in-glass thermometer are laboratory
thermometer and clinical thermometer.
Liquids used in liquid-in-glass thermometers
There are two major liquids used in liquid thermometers. These liquids are mercury and Ethanol
(alcohol). Each of these liquids has the following advantages and disadvantages:
Mercury
The advantages of using mercury in a liquid thermometer are:
a. It expands uniformly.
b. It does not wet the sides of the tube or it does not cling to the walls of the tube.
c. It is a good conductor of heat.
d. It has a high boiling point of 357
o
C.
e. Its specific heat capacity is very low.
The disadvantages of using mercury in liquid thermometers are:
a. It freezes at -39
o
C, therefore mercury cannot be used in very cold regions that have
temperatures below -39
o
C.
b. It is poisonous. It would cause health hazards if the tube broke.
c. Its expansivity is low.
d. It is very expensive.
Alcohol
The advantages of using alcohol in liquid thermometers are:
a. It expands uniformly and its freezing point is -115
o
C. Therefore, it can be used in very
cold regions.
b. It has a large expansivity. Alcohol can therefore be used in wide tubes as well.
The disadvantages of using alcohol in liquid thermometers are:
a. It has to be coloured to be seen clearly.
b. It wets the tube because it clings to the walls of the tube.
c. It has a low boiling point of 78
o
C.
d. Its thread has a tendency of breaking.
e. It has a high specific heat capacity.
56
Clinical thermometer
A clinical thermometer is used to take the temperature of the body. The thermometer uses
expansion and contraction of mercury. It has a constriction in the capillary tube just above the
bulb.
How a clinical thermometer works:
When a clinical thermometer is put in the patient’s mouth or under the armpit the temperature
rises. The mercury in the capillary tube expands and it is pushed through the constriction and up
the tube.
When a clinical thermometer is taken out of the patient’s mouth or armpit, the mercury cools and
contracts. The mercury cannot go back through the constriction and the thread breaks. This is an
advantage because the mercury in the tube cannot go back into the bulb and patient’s
temperature can be read off. You must shake the thermometer to get the mercury back into the
bulb.
The scale of the clinical thermometer ranges from 35
0
C to 42
0
C since the normal body
temperature is only 37
0
C. The short range enables the thermometer to be short. The thermometer
is more accurate and has high sensitivity.
Figure 4.2 clinical thermometer
Bulb
Constriction
Very thin bore
Scale
Expansion
chamber
57
Thermocouple thermometer
A thermocouple thermometer is a thermometer which uses a thermo-electric property. It consists
of two wires of different materials e.g. copper and iron.
How a thermocouple thermometer works
When the hot junction is heated an electric current flows and produces a reading on a sensitive
meter. The value of current produced depends on the temperature difference.
Advantages of a thermo couple thermometer
a. It has low heat capacity. It can be used to measure fluctuating temperatures.
b. It has a very large range. The range is from -200
0
C to 1500
0
C.
c. It can measure the temperature at a point.
Disadvantage: A thermocouple thermometer can only be used over a certain temperature range
where variation of current with temperature is uniform.
Resistance thermometer
Resistance thermometer uses the variation with temperature of the resistance of a coil of wire.
For example: In platinum wire, the resistance of the wire decreases with an increase in
temperature. In nichrome wire, the resistance of the wire increases with an increase in
temperature.
Figure 4.3: thermocouple
58
Advantages of a resistance thermometer
a. It is far more accurate.
b. It has a very large range.
c. It can be read at a distance if it has longer leads. This enables the observer to be far from
where the temperature is being measured, e.g. in a blast furnace.
Summary
The two temperature scales are Celsius scale and Kelvin scale.
To convert Celsius to Kelvin you add 273 to temperature in degrees Celsius (K =
0
C +273).
To convert Kelvin to Celsius you subtract 273 from temperature in Kelvin (
0
C = K-273).
Temperature is measured by instruments called thermometers. Thermometers measure
temperature by using physical properties.
Various types of thermometers are:
Liquid-in-glass thermometer, e.g. laboratory thermometer and clinical thermometer
Thermocouple thermometer
Resistance thermometer
Student assessment
1. Define
a. Temperature
b. Heat
Exercise 4.2
In your groups, answer the following questions:
1. Explain why
a. mercury is used in thermometers
b. a clinical thermometer has a constriction.
2. Discuss how you could check the lower fixed point and the upper fixed
point of a liquid-in-glass thermometer.
3. With the aid of a diagram, describe how a thermocouple thermometer
works.
59
2. Discuss
a. Lower fixed point of a liquid in glass thermometer.
b. Upper fixed point of a liquid in glass thermometer.
3. 200
0
C 100
0
C 150
0
C 0
0
C 20
0
C 37
0
C -273
0
C 78
0
C -100
0
C 25
0
C
From the above list of temperatures choose the most likely value for each of the
following:
a. The room temperature
b. The normal body temperature
c. The melting point of water
d. The boiling point of water
e. The freezing point of water
f. The boiling point of ethanol
g. Absolute zero
4. With the aid of a well labeled diagram explain how a clinical thermometer works.
5. Sate three physical properties that are used by thermometers to measure temperature.
6. Convert the following into Kelvin
a. -250
0
C
b. 47
0
C
c. 110
0
C
7. Convert the following into degrees Celsius
a. 550K
b. 470K
c. 10K
8. Give two differences between a laboratory thermometer and a clinical thermometer.
9. Explain why an alcohol liquid thermometer might be preferred to a mercury liquid
thermometer in the arctic region.
10. List the two advantages and two disadvantages of using mercury as thermometer liquid.
11. List two advantages and two disadvantages of using alcohol as thermometer liquids.
12. Water is unsuitable for use in thermometers. State a reason for this.
60
13. Explain why a liquid-in-glass thermometer has the following:
a. A thin walled glass bulb.
b. A thick glass wall.
c. A very narrow capillary tube.
14. If you want to manufacture a new liquid-in-glass thermometer, describe the changes that
could be made to:
a. give the thermometer a greater range
b. make the thermometer more sensitive.
15. With the aid of a well labeled diagram, explain how a thermocouple works.
61
Pressure
Objectives
At the end of chapter 5, you must be able to:
Define pressure
Determine pressure exerted by regular solids
Describe experiments to investigate factors
affecting pressure in liquids
5.1 What is pressure?
When forces act on a surface their effect is spread over an area. This effect creates pressure.
Pressure is defined as the force exerted per unit area.
Pressure is calculated by dividing the force acting at right angles to the surface by the area over
which it acts.
Pressure = Force
Area on which force acts
P = F
A
If force is measured in Newtons (N) and area in cm
2
, then pressure is measured in N/cm
2
.
If force is measured in Newtons (N) and area in m
2
, then pressure is measured in N/m
2
.
1N/m
2
is equivalent to 1 Pascal.
1N/m
2
= 1 Pa
For example:
If a force of 50 N acts on an area of 10cm
2
, the pressure is 5 N/cm
2
If a force of 50 N acts on an area of 10m
2
, the pressure is 5 Pa.
Chapter 5
F
P
A
62
5.2 Pressure exerted by solids
Figure 5.1 shows a box in position X and position Y.
In position X a box is exerting a force on a smaller area while in position Y a box is exerting a
force on a larger area. When the force is spread over a larger area, pressure is reduced because
the force on each square metre is reduced and vice versa. From Figure 5.1, the pressure under
block X is less than the pressure under block Y.
Factors that affect pressure exerted by solids
1. Contact surface area
As it was explained in Figure 5.1, the size of pressure is affected by the surface area on which
force is exerted.
A large surface area causes less pressure. A small area of contact increases pressure or causes
high pressure. This can be demonstrated by the following examples:
Weight = 16 N
5m 4m
2m
2m 5m
4m
Figure 5.1
Area = 4m x 2m = 8m
2
Area = 5m x 2m = 10 m
2
pressure = 16N pressure = 16N
8m
2
10m
2
Pressure = 2Pa Pressure = 1.6 Pa
X
Y
63
a. Figure 5.2 shows two wheels of two cars of a farmer.
Figure 5.2 wheels of cars
When the ground is very soft, a farmer is encouraged to use a vehicle with wheel B because it
has a large flat surface. A large surface produces less pressure to the ground.
If a farmer uses a vehicle with wheel A, the car is likely to sink due to the small area of contact.
This causes high pressure.
b. If you stepped on the point of a sharp nail with your bare foot, it would be extremely
painful because the surface area is very small. Hence a sharp nail exerts greater pressure
on your bare foot.
c. If you lie on a bed of nails-points with a large number of nails, it would not be extremely
painful because the surface area of the nails has increased. Hence sharp-nails exert less
pressure on your body.
2. Size of the force
Pressure exerted by solids can increase with an increase in the size of the force when surface area
is kept constant because more force acts per given area.
smaller surface area
Large surface area
64
Worked examples
1. A block weighing 200 N rests on an area of 2 m
2
. Calculate the pressure exerted by the
block on the surface which supports it.
Solution
F =200 N A= 2 m
2
P =?
P = F
A
P= 200 N
2 m
2
P= 300 N/m
2
OR P = 300 Pa.
2. Pressure exerted by a regular solid of base area 10cm
2
is 3N/cm
2
. Calculate the weight of
a solid.
Solution
P=3N/cm
2
A= 10 cm
2
F=?
F = P x A
F = 3 N/cm
2
x 10 cm
2
F = 30 N
3. A block of mass 20 kg has the base measured 0.2 m x 1.5 m. Calculate the pressure
exerted by the block to the ground.
Solution
F = 20 x 10 N = 200 N A= 0.2 x 1.5 = 0.3 m
2
p =?
P= F
A
P = 200 N
0.3 m
2
P= 666.7 Pa
65
5.3 Pressure in liquids
Pressure in liquids is caused by the force exerted by the liquid molecules on the wall of the
container.
Internal stresses are set up in the liquid by external forces, and these allow the pressure in a
liquid to be transmitted in all directions.
Factors affecting pressure in liquids
1. Pressure in a liquid increases with depth
The deeper into the liquid, the greater the pressure because as you go deep the weight of the
liquid above increases. When the weight of the liquid increases, pressure also increases.
Exercise 5.1
In your groups, answer the following questions:
1. Explain why
a. the area under the edge of the knife is extremely small
b. wall foundations have a large horizontal area.
2. The pressure exerted by the solid to the ground is 50 Pa. What does this mean?
3. A force of 100 N acts on an area of 4 m
2
.
a. Calculate the pressure produced.
b. What would the pressure be if
(i) The area is halved?
(ii) The area is doubled?
4. The rectangular block of mass 10 kg has measurements 0.2 m by 0.5 m by 0.3 m.
Calculate the pressure produced by the block.
5. What force is produced if a pressure of 200 Pa acts on an area of 0.5 m
2
?
66
RESULTS
Water from outlet C is thrust further horizontally compared to outlet B. Water from outlet B is
thrust further horizontally compared to outlet A. Beaker catching water from outlet C has the
highest level of water, seconded by beaker catching water from outlet B then A has the lowest
level of water.
Experiment 5.1
AIM: To show that pressure in a liquid increases with depth.
MATERIALS: Spouting can, 3 beakers and water
PROCEDURE:
1. Punch 3 equal sized holes at different heights in a spouting can as shown in
Figure 5.3 below.
2. Stand the spouting can on one end.
3. Position 3 beakers to catch the water.
4. Fill the spouting can with water.
5. Keep on adding water in the spouting can for some minutes.
6. From which outlet is water thrust further horizontally?
7. Which beaker has highest water level?
Figure 5.3
A B C
A
B
C
67
EXPLANATION
Water squirts (comes out) with greatest pressure at outlet C.
Water squirts (comes out) with least pressure at outlet A.
CONCLUSION
This shows that the pressure of water is greatest at the deepest point in the liquid. Therefore,
pressure in liquid increases with depth.
2. Density
Pressure in liquid increases with density because more dense liquid is heavier or has a greater
weight than a less dense liquid of the same volume. For example, if 1l of water of density
1g/cm
3
and 1l of mercury of density 13.6g/cm
3
are placed in identical containers, mercury will
produce more pressure at the bottom of the container than water would because mercury is more
dense than water.
68
RESULTS:
When a solid was immersed in water the apparent loss in weight was less than the apparent loss
of weight when it was immersed in mercury.
EXPLANATION/CONCLUSION
Apparent loss of weight in water was less because water is less dense and has less weight.
Apparent loss of weight in mercury was greater because mercury is denser and has more weight.
The pressure exerted by water on a given area of a solid is less than the pressure exerted by
mercury. Therefore, pressure in liquid is affected by density.
Experiment 5.2
AIM: To show that pressure in liquid is affected by density
MATERIALS: A solid, spring balance, water, mercury, 2 beakers and ruler.
PROCEDURE:
1. Pour the same volume of water and mercury in separate identical beakers.
2. Weigh the block in air using the spring and record its weight as W1.
3. Immerse the base of the solid in water to a measured height.
4. Record the new weight on the spring as W2.
5. Calculate the weight of water as:
Weight of water = apparent loss in weight
Weight of the water = W2- W1
6. Repeat the experiment by immersing the block of wood in mercury to the same
measured height.
Figure 5.7
W2
W1
69
Calculation of pressure in a liquid
The pressure of a liquid depends on its density and depth.
To derive the formula p = ρhg
Using the container shown in Figure 5.8 below:
We consider base area of a liquid, A at a depth (height) of h and density of a liquid ρ.
Volume of a liquid = base area x depth = Ah
Mass of a liquid = density x volume = ρ x v
But v = A x h
Therefore, mass of a liquid = ρAh
Weight of a liquid = mass x acceleration due to gravity = m x g
But mass = ρAh
Therefore, weight of a liquid = ρAhg
Pressure = Force (weight)
Area
Pressure = ρAhg
A
Therefore, pressure = ρhg.
The pressure of a liquid = density x depth (height) x acceleration due to gravity
Figure 5 .6 Pressure in liquid
70
Whereby: density is in kg/m
3
, depth(height) is in m and acceleration due to gravity is 10 m/s
2
.
Worked examples
1. Calculate the pressure exerted by a column of a liquid at the base of a container if the
density of a liquid is 13600 kg/m
3
and its depth is 0.1m. (g = 10m/s
2
)
solution
ρ = 13600kgm
-3
h= 0.1m p =? g = 10m/s
2
P =ρhg
P = 13600 kg/m
3
x 0.1 x 10 m/s
2
P = 13600 pa OR P= 13.6 Kpa
2. A pressure of 1000pa is exerted by a column of petrol in a tank of a car. Calculate the
height of the petrol column (Density of petrol = 800kg/m
3
, g = 10m/s
2
)
P =1000pa ρ = 800kgm
-3
g =10ms
-2
h =?
P = ρhg
h = p
ρg
h = 1000 Pa
800kg/m
3
x 10m/s
2
H = 0.125m OR h= 12.5 cm.
5.4 Pascal’s principle
Blaise Pascal was a French mathematician. Pascal came up with his principle of transmission of
fluid pressure. Pascal’s principle of transmission of pressure in fluids states that pressure exerted
anywhere in an enclosed incompressible fluid is transmitted equally in all directions throughout
the fluid.
The pressure applied anywhere to a body of fluid causes a force to be transmitted equally in all
directions; the force acts at right angles to any surface in contact with the fluid. This causes the
pressure variations (initial differences) to remain the same.
71
DISCUSSION
When the piston on the plunger on the large syringe is pushed in, the person holding the plunger
on the smaller syringe will feel the plunger moving out. The pressure has been transmitted
through the liquid in the system.
Using Pascal’s principle, the pressure in a large syringe equals the pressure in the smaller
syringe.
Pressure in a large syringe = pressure in a small syringe
Force = Force
Area for the large syringe Area for the small syringe
Experiment 5.3
AIM: To investigate the transmission of pressure in liquids
MATERIALS: Large syringe, smalle syringe, water and a pipe.
PROCEDURE:
1. Set up the experiment as shown in Figure 5.7 below.
2. Push the plunger on the larger syringe while holding the plunger on the smaller syringe.
How do you feel on the smaller syringe?
Large syringe
Small syringe
Force
Figure 5.7
72
Force = Force
larger area small area
Worked example
If the force on a large syringe is 80 N and the area is 0.5 m
2
, calculate the force on the small
syringe with area 0.1 m
2
.
Solution
Pressure in a large syringe = pressure in a small syringe
Force = Force
large area small area
80 N = Force on a small syringe
0.5 m
2
0.1 m
2
Force on the small syringe = 80N x 0.1 m
2
0.5 m
2
Force on a small syringe = 16 N
5. 5 Atmospheric pressure
Atmospheric pressure is experienced because air that is called atmosphere exerts pressure on
objects. This pressure is normally called Air pressure.
Exercise 5.2
In your groups, answer the following questions:
1. Calculate the pressure exerted by a column of water at the base of a
container if the density of water is 1000 kg/m
3
and its depth is 1m.
(g = 10m/s
2
)
2. A pressure of 60 000 pa is exerted by a column of mercury in a
container. Calculate the height of the mercury column
(Density of petrol = 13600 kg/m
3
, g = 10m/s
2
)
3. Discuss Pascal’s principle of transmission of pressure in fluids.
73
Demonstrating atmospheric pressure
1. Collapsing can experiment
The steam produced in a can replaces the air and molecules in the steam and exerts pressure on
the walls of the can equal to the atmospheric pressure. The tightly screwed can suddenly
crumples when it is cooled by running cold water. The steam inside the can has condensed into a
very small volume of water. This leaves a partial vacuum behind. The decrease in temperature
decreases the kinetic energy of molecules inside the can. The pressure inside the can decreases
Experiment 5.4
AIM: To demonstrate atmospheric pressure
MATERIALS: A can, tap water, very cold water, heat source
PROCEDURE:
1. Pour some tap water into a can.
2. Heat some water in an open can until it boils.
3. Remove the can from the heat and screw on its cap tightly.
4. Cool the can by running cold water over it.
The can suddenly crumples as shown in Figure 5.8 below. Explain why.
Figure 5.8 crushing can experiment
74
and it is less than the atmospheric pressure. The pressure difference on opposite sides of the
walls of the can results in a very large unbalanced force acting inwards.
2. Drinking straws
In the drinking straw, air is first sucked out of the straw. The pressure of air inside the straw is
less than the atmospheric pressure which is pressing down on the surface of the liquid outside the
straw. Therefore, the liquid is forced out up the straw and into the mouth.
3. Rubber sucker
When the moistened concave surface of the rubber sucker is pressed against a flat surface the air
between the two surfaces is squeezed out. This leaves the pressure in the enclosed space much
reduced and creates a vacuum. The atmospheric pressure acting on the sucker forces the sucker
against the flat surface.
4. Vacuum cleaner
In a vacuum cleaner, a fan lowers the air pressure just beyond the bag. This creates a pressure
difference between the inside and outside the vacuum cleaner. The atmospheric pressure rushes
Figure 5.9 drinking through a straw
Figure 5.10 a sucker
75
in, carrying dust and dirt with it. The dust and dirt is stopped by the bag but the air is not
stopped.
Measuring atmospheric pressure
The instruments that are used to measure atmospheric pressure are called barometers.
The following are the types of barometers used:
1. Mercury barometer
A simple mercury barometer uses a thick walled tube of about 1 m long. It uses mercury that is
poured into the tube. This mercury tube is inserted into a wider vessel containing mercury. Some
mercury runs out of the tube into the vessel leaving the space at the top of the tube. The space
left at the top of the tube is called a Vacuum.
Figure 5.11 vacuum cleaner
76
How a mercury barometer works
When atmospheric pressure is acting on mercury in the vessel it pushes mercury downwards and
forces it up into the tube. The mercury rises to a height that is equivalent to atmospheric
pressure. The height of mercury, h is measured on a ruler.
Standard pressure
At sea-level, the mercury level inside the tube rises up to 760 mm of mercury (760 mm Hg). The
height, h of mercury at sea-level is called Standard atmospheric pressure. Therefore, standard
atmospheric pressure is 760 mmHg. 760mm Hg can also be expressed as 0.76 m Hg or
76 cm Hg.
Worked example
Calculate the atmospheric pressure in Pascals (pa) when a mercury barometer supports a column
of mercury 76 cm high. (Density of mercury = 13600kg/m
3
)
Solution
P =? h = 76 cm = 0.76 m ρ = 13600kg/m
3
g = 10m/s
2
P = ρhg
P = 13600kg/m
3
x 0.76m x 10m/s
2
Pressure = 103360pa.
2. Water barometer
Water can be used in a barometer instead of mercury. The first water barometer was built by
Von Guericke in the seventh century. A longer tube is required to use water in a barometer
because water is far much less dense than mercury. For example, at sea level where the
Figure 5.12 diagram of a simple mercury barometer.
77
atmospheric pressure is 100 000 Pa, we can calculate the height of the water in the tube as
follows:
density of water = 1000 kg/m
3
, g = 10m/s
2
and p = 100 000 Pa
pressure = ρgh
100 000 pa = 1000 kg/m
3
x 10 m/s
2
x h
h = 100 000 pa
1000 kg/m
3
x 10 m/s
2
h = 10 m
Therefore, the level of the column of water at sea level is 10 m. The water barometer is not very
useful in practice.
3. Aneroid barometer
The main feature of the aneroid barometer is the small sealed metal box containing air at low
pressure.
How an aneroid barometer works.
Atmospheric pressure tries to squash the metal box which is corrugated to make it more flexible
in the middle. If the pressure rises, the top and the bottom of the metal box become even more
squashed in. The movement of the box is magnified by a lever. The lever pulls a chain which
moves the pointer further up the scale.
Aneroid barometers are more portable, much easier to use and cheaper than mercury barometers.
Figure 5.13 aneroid barometer
78
5.6 Applications of pressure
Applications of pressure in liquids
Pressure in liquids is used in the following:
1. Hydraulic machines
Hydraulic machines are used to lift the weight of a body. Examples of hydraulic machines are
hydraulic folk lifters, hydraulic jacks, hydraulic brakes and hydraulic loaders. Hydraulic
machines operate by using Pascal’s principle of transmission of pressure in fluids. This happens
because liquids are incompressible, so when the liquid is pressed, pressure is transmitted to all
parts of the liquid and the pressure is the same.
In hydraulic machines a small force (effort) move a large force (load) as shown in Figure 5.14.
Exercise 5.3
In your groups, answer the following questions:
1. Calculate the value of atmospheric pressure in Pa when the level of mercury in
a mercury barometer is 700 mm Hg.
2. Explain why mercury is used in a barometer rather than water.
3. Thee Atmospheric pressure is 103333 Pa. Calculate the height of mercury in
the tube in:
a. mm Hg b. m Hg c. cm Hg
4. State two common properties of atmospheric pressure and liquid pressure.
79
When an effort, F
1
is applied at A
1
, the small piston is pushed down and a liquid gets pushed
through the pipe.
The liquid is forced to push the load at A
2
upwards. The same pressure applied by the liquid at
A1 is the same pressure that is used to lift up the load.
Pressure at A
1
= force
Area
Pressure at A
2
= pressure at A
1
Upward force on a load at A
2
= pressure at A
1
x Area at A
2
Worked example
Figure 5.15 is a diagram showing a simple hydraulic machine used to lift a load.
a. Calculate the pressure exerted on the liquid by an effort of 40N.
b. Calculate the thrust (Force) on the load.
Figure 5.14 hydraulic machine.
Figure 5.15
A1=0.02 m
2
A2 = 3 m
2
80
Solution
Effort = 40 N A
1
= 0.02 m
2
Load =? A
2
= 3m
2
a. Pressure exerted by a load of 40N
P = F = effort
A A
1
P = 40 N
0.02 m
2
P = 2000 pa
b. Thrust(force) on the load
Force = pressure x area, A
2
F = 2000 pa x 3 m
2
F = 6000 N.
OR
Force (effort) = Force (load)
A
1
A
2
Force (load) = Force (effort)
A2 A
1
Force (load) = Force (effort) x A2
A
1
= 40 N x 3 m
2
0.02 m
2
Force on the load = 6000N.
2. Construction of dams
Liquid pressure is used in construction of dams. The bottom of the dam is made thicker in order
to withstand liquid pressure which increases with depth.
81
3. Water supply systems
The water supply comes from a reservoir on high ground. The water flows through the pipe to
the taps and storage tanks that are at a lower level because liquid pressure increases with depth.
Applications of Atmospheric (air) pressure
The following are the applications of atmospheric pressure in our everyday life:
1. Drinking straw
2. Rubber sucker
3. Syringe
These applications have already been explained in section 5.5.
5.7 Archimedes’ principle
Why do ships float on water when in fact they should sink? Why does paper float on water and a paperweight
sink?
The underlying fact on which the principle is based was discovered by Archimedes in about 300 B.C. The
story goes that the King of Sicily suspected that the goldsmith had mixed some silver in his crown and
cheated him. Without destroying the crown, he wanted to know the truth. Archimedes was asked to find out
whether this was so, without destroying the crown.
One day, while getting into his bath he noticed water spilling over the sides. In a flash, Archimedes realised
the relationship between the water that had splashed out and the weight of his body. It seems that
Archimedes got the solution. Archimedes was so excited with his discovery that he leapt out of the bath, and
rushed naked into the streets yelling triumphantly, 'Eureka!' 'Eureka!' (Greek word for 'I have found it!).
He obtained a lump of pure gold and a lump of pure silver, each with a weight equal to that of the crown. By
immersing each in a vessel full of water he collected the volume of water which overflowed. The volumes
were all found to be different. So the crown was not pure.
82
Archimedes’ principle
When an object is immersed in a liquid the liquid exerts an upward force which is known as
upthrust. Therefore, an object weighs less in water than in air. If an object weighs 10 N in air
and it weighs 7 N when immersed in water, the upthrust is found as:
Upthrust = Weight of an object in air– weight of an object in water
Upthrust = 10 N – 7 N = 3 N
The weight of an object in water which is lower than the real weight of an object is called
apparent weight.
Upthrust is proportional to the weight of displaced liquid.
The upthrust of the object in Figure 5.16 = 0.67 N – 0.40 N = 0.27 N
The 0.67 N object has an upthrust of 0.27 N. Therefore, it displaces 0.27 N of water.
The 0.67 N object feels like it only weighs 0.40 N under water. Therefore, apparent weight of an
object is 0.40 N.
Archimedes’ principle states that ‘if a body is totally or partially immersed in a fluid (gas or
liquid) the fluid exerts an upthrust which is equal to the weight of the fluid displaced’.
Figure 5.16 apparent weight
83
Experiment 5.5
AIM: To verify Archimedes’ principle
MATERIALS: Beaker, mass, displacement can, spring balance, top pan balance, a small
block
PROCEDURE:
1. Suspend (hang) the block in air from the spring balance. Record its weight.
2. Fill the displacement can with water until it reaches a point of overflowing.
3. Place a clean, dry empty beaker on a top pan balance. Calculate its weight as m
(in kg) x 10 kg/N.
4. Place an empty beaker under the spout of the displacement can.
5. Carefully lower the block, still attached to the spring balance.
6. Record the new weight of a block.
7. Calculate the apparent loss in weight (upthrust).
8. Place a beaker containing displaced water on a top pan balance. Calculate its
weight as m (in kg) x 10 kg/N.
9. Calculate the weight of the displaced water as follows:
Weight of displaced water = weight of beaker containing water – weight of an
empty beaker.
10. Set down the results as follows:
Weight of a block in air = N
Weight of a block in water = N
Apparent loss in weight of a block (upthrust) = N
Weight of an empty beaker = N
Weight of a beaker with displaced water = N
Weight of displaced water = N
Compare the apparent loss in weight of a block and weight of displaced water.
Discuss your results with your friends in class.
Figure 5.16
84
EXPLANATION/CONCLUSION
From experiment 5.5 the apparent loss in weight (upthrust) of a block must be equal (or
approximately equal) to the weight of displaced water.
Apparent loss in weight of a block (upthrust) = weight of the displaced water.
The displacement can in Archimedes’ principle is also called Eureka can.
Relative density of a substance
Relative density of a substance is the ratio of the density of any volume of substance to the
density of an equal volume of water.
Relative density = density of any volume of substance
density of an equal volume of water
For example: If the density of mercury is 13600 kg/m
3
and the density of equal volume of water
is 100 kg/m
3
,
Relative density of mercury = 13600 kg/m
3
= 13.6
1000 kg/m
3
Relative density of a substance can also be defined as the ratio of the mass of any volume of
substance to the mass of an equal volume of water.
Relative density = mass of any volume of substance
mass of an equal volume of water
For example: If the mass of a block is 10 kg and the mass of water is 1 kg,
Relative density of a block = 10 kg = 1
1 kg
Relative density of a substance can also be defined as the ratio of the weight of an object to the
apparent loss of weight in water.
Relative density = weight of an object
apparent loss of weight in water
For example: If the weight of a block is 100 N and the apparent loss of weight in water is 40 N,
Relative density of mercury = 100 N = 2.5
40 N
85
Law of floatation
An object which is placed in a fluid will float if the upthrust acting on it is strong enough to
support its weight.
If a 100 N block is lowered into water, the upthrust acting on it rises. This causes more water to
be displaced. The block will float when the upthrust reaches 100 N. Therefore, the weight of the
displaced water is also 100 N. This means that the weight of the displaced water is equal to the
weight of a block.
Weight of a displaced water = upthrust = weight of a block
The law of floatation states that a floating object displaces its own weight of the fluid in which
it floats.
The concept of the law of floatation can be applied when considering why objects float.
Consider 100 000 N block of solid iron. As iron is nearly eight times denser than water, it
displaces only 1/8 of 100 000 N of water when submerged, which is not enough to keep it afloat.
Suppose the same iron block is reshaped into a ship. It still weighs 100 000 N, but when it is put
in water, it displaces a greater volume of water than when it was a block. The deeper the iron
ship is immersed, the more water it displaces, and the greater the upthrust force acting on it.
When the upthrust force equals 100 000 N, it will sink no farther.
Since a floating object displaces a weight of fluid equal to its own weight, every ship must be
designed to displace a weight of fluid equal to its own weight. A 100 000 N ship must be built
wide enough to displace 100 000 N of water before it sinks too deep in the water. The same is
true for vessels in air (as air is a fluid): an aeroplane that weighs 10 000N displaces at least
10 000 N of air. If it displaces more, it rises; if it displaces less, it falls. If an aeroplane displaces
exactly its weight, it hovers at a constant altitude.
Applications of Archimedes’ principle
Archimedes’ principle and law of floatation can be used in:
1. Floating of solids in a fluid: When a block is dropped in water, it floats because it
displaces water equal to its own weight. The density of a block of solid is less than the
density of water. A large ship floats on water by using law of floatation. This is also
possible because the ship contains a lot of spaces that are filled with air. The average
density of the ship becomes less than the density of water.
2. Explaining why hot-air balloons rise: When the air in the balloon increases, its volume
also increases. An increase in volume increases the weight of the air displaced by the
balloon. The balloon then floats. The balloon can also float when the gas burner can heat
86
the air inside at 100
0
C. The air in the balloon expands and pushes out through the hole at
the bottom. This reduces the weight of the air in the balloon. The hot air also becomes
less dense which helps in the floating of the hot air balloon.
3. Checking the purity of a material as it was done by Archimedes.
4. Hydrometer: a useful instrument in which the Principle of floatation is applied. It floats
at different levels in liquids of different densities.
Hydrometer floats less in methylated spirit than water because methylated spirit is less
dense than water.
The hydrometer sinks in the liquid and only floats when the weight of the liquid
displaced is equal to the weight of the hydrometer. Therefore, it is used to measure the
density of the liquid in kg/m
3
, check the quality of beer and milk and test the state of
charge of car batteries.
Calculations on Archimedes’ principle and floatation
Worked examples
1. A boat floating on water weighs 8 000 N.
What is
a. the upthrust acting on the boat?
b. the weight of the water displaced by the boat?
Solution
a. weight of a boat = upthrust = 8 000 N
b. weight of displaced water = weight of a boat = 8000 N
2. The mass of a metal bar in air is 0.5 kg and its mass is 0.3 kg when immersed in water.
Calculate;
a. The weight of a metal bar in air.
b. The weight of a metal bar when immersed in water.
c. The apparent loss in weight of a metal bar.
d. The upthrust acting on a metal bar.
Solution
a. Weight in air = m x g
= 0.5 kg x 10 m/s
2
Weight in air = 5 N
b. Weight when immersed in water = m x g
= 0.3 kg x 10 m/s
2
Weight when immersed in water = 3 N
87
c. Apparent loss in weight = weight in air – weight when immersed in water
= 5 N – 3 N
Apparent loss in weight = 2 N
d. Upthust = apparent loss in weight = 2N
Summary
Pressure is the force exerted per given area. Pressure is measured in Pascals.
Pressure exerted by a regular solid depends on the size of force and its surface area.
Properties of liquid pressure are:
it acts in all directions
depends on density
depends on depth
Pressure in a liquid is found by using a formula, P = ρgh
Pascal’s principle in liquids states that pressure exerted anywhere in an enclosed incompressible
fluid is transmitted equally in all directions throughout the fluid. Pascal’s principle in liquids is
applied in hydraulic machines, syringes and siphons.
Exercise 5.4
In your groups, answer the following questions:
1. Explain the following:
a. A hot air balloon floats when the air inside is heated by a burner.
b. A steel ship floats on water.
2. A block of metal of mass 2000 kg floats on water.
a. Calculate the size of upthrust acting on a block of metal.
b. State the weight of the displaced water.
3. The density of petrol is 800 kg/m
3
and the density of water is 1000 kg/m
3
. In which
liquid will the object experience more upthrust? Give a reason for your answer.
4. An iron bar weighs 500 N in air and it weighs 200 N when immersed in paraffin.
Calculate
a. The apparent loss in weight of a metal bar
b. The upthust acting on an iron bar.
88
Atmospheric pressure is caused by the force exerted by air particles per given area on the walls
of the object in the atmosphere. Atmospheric pressure is used in drinking straws, rubber suckers
and vacuum cleaners. Atmospheric pressure is measured by an instrument called barometer, e.g.
mercury barometer.
Archimedes’ principle states that ‘if a body is totally or partially immersed in a fluid (gas or
liquid) the fluid exerts an upthrust which is equal to the weight of the fluid displaced’.
The law of floatation states that a floating object displaces its own weight of the fluid in which
it floats.
Archimedes’ principle and law of floatation is applied in floating solids on a liquid, floating of a
hot air balloon and hydrometers.
Student assessment
1. Define
a. pressure
b. upthrust
2. With the aid of a diagram explain how atmospheric pressure is caused.
3. State three factors that affect pressure in liquids.
4. A block weighing 200 N rests on an area of 2 m
2
. Calculate the pressure exerted by the
block on the surface which supports it.
5. Derive a formula to show that the pressure of a liquid depends on depth and density.
6. At atmospheric pressure a barometer reads 76 cm. If one atmosphere equals to
101 000 Pa, calculate the density of mercury.
7. Calculate the pressure exerted by a column of a liquid at the base of a container if the
density of a liquid is 13600 kg/m
3
and its depth is 0.1 m. (g = 10 m/s
2
).
8. A pressure of 1000 pa is exerted by a column of petrol in a tank of a car. Calculate the
height of the petrol column. (Density of petrol = 800 kg/m
3
, g= 10 m/s
2
).
89
9. Figure 5.17 is a diagram showing a simple hydraulic machine used to lift a load.
a. Calculate the pressure exerted on the liquid by an effort of 40N.
b. Calculate the thrust (Force), F2.
10. Calculate the atmospheric pressure in Pa when a mercury barometer supports a column of
mercury 76 cm high. (Density of mercury = 13600 kg/m
3
)
11. The mass hits a wall. The average force exerted on the wall during the impact is 100 N.
The area of the mass in contact with the wall at impact is 0.03 m
2
. Calculate the average
pressure that the mass exerts on the wall during the impact.
12. a. A woman weighs 700 N and the total area of her shoes in contact with the ground is
0.0025 m
2
. A man weighs 600 N and the total area of his shoes in contact with the round
is 0.003 m
2
. Calculate the pressure exerted by
i. the woman
ii. the man to the ground.
b. Which one exerts greater pressure on the ground?
13. State
a. the Archimedes’ principle
b. the law of floatation.
14. A bowl weighs 10 N. If it floats in water, state
a. the upthrust exerted on the boil
b. the weight of the displaced water.
Figure 5.17
40
N
A=20 cm
2
A=50 cm
2
90
15. The density of water is 1000 kg/m
3
. Use table 5.1 to answer the questions that follow.
Table 5.1
a. Which substance (s) will float in water?
b. Give a reason for your answer in 15 (a).
c. For substance (s) in 15 (a), state the upthrust exerted on each substance by water.
16. A loaded tanker weighs 3 300 000 N and it floats on the lake.
a. State
i. the size of upthrust on the loaded tank
ii. the weight of the water displaced by the loaded tanker.
b. If 500 000 N of oil is off loaded,
i. Calculate the weight of the loaded tanker.
ii. State the size of the upthrust on the loaded tanker.
c. Explain what happens on the tanker after off loading the 500 000 N of oil.
17. A block of solid weighs 25 N in air and it weighs 15 N when immersed in water.
a. Calculate the apparent loss in weight.
b. State the upthrust exerted on the block of solid by water.
c. State the weight of the displaced water.
d. What will happen to the apparent loss in water if
i. a block was immersed in a liquid less dense than water?
ii. a block was immersed in a liquid denser than water?
18. The depth of the water in a dam is 4 m.
a. Calculate the pressure exerted by water at the bottom of the dam. (g = 10m/s
2
and density
of water = 1000 kg/m
3
)
b. Explain why the dam is thicker at the base than at the top.
Substance
Density (kg/m
3
)
Mercury
13600
Paraffin
800
Polythene
950
Pine
500
Granite
2700
91
19. Figure 5.18 shows a pipe with one end wider than the other, containing oil.
a. Calculate the pressure applied to the oil if piston A is pushed into the pipe.
b. What pressure is exerted at piston B?
c. Calculate the force exerted at piston B.
d. Explain how Pascal’s principle of transmission of pressure in fluids is applied.
20. With the aid of a well labeled diagram, explain how a mercury barometer is used to measure
atmospheric pressure.
21. Describe an experiment that you would carry out in order to verify Archimedes’ principle.
Figure 5.18
Piston A
Force = 20 N
Area = 10 cm
2
Oil
Piston B
Area = 40 cm
2
92
Gas laws
Objectives
After the end of chapter 6, you must be able to:
Discuss gas laws
Explain applications of gas laws
6.1 Gas laws
Behaviour of a gas depends on three factors:
Pressure
Volume
Temperature
Gas laws were discovered by using the volume, temperature and pressure of the gas.
There were relations that were used to describe the gas laws. In these relations two of the factors
mentioned above were varied while one factor was kept constant.
Chapter 6
93
Boyle’s law
Boyle was relating pressure and volume at constant temperature. Boyle’s law was published in
1662.
Experiment 6.1
AIM: To investigate the relationship between pressure and volume at constant temperature.
MATERIALS: Boyle’s law apparatus, foot pump
PROCEDURE:
1. Connect the apparatus as shown in Figure 6.1.
2. Pump in air by using a foot pump.
3. Check and record the volume of the trapped air in the tube then check and record
the pressure reading on the Bourdon pressure gauge.
Explain what you have noticed.
Figure 6.1
Trapped air
Oil
Bourdon pressure
gauge
Air from foot
pump
94
RESULTS
When the air is pumped in from the foot pump, the level of air in the reservoir decreases and the
oil is pushed upwards in the glass tube.
The following will be noticed:
The volume of the trapped air in the tube decreases.
The reading on the pressure gauge increases.
The readings can be plotted on the graph as shown in Figure 6.2.
EXPLANATION
From the graph in Figure 6.2, we can notice that if the volume halves, the pressure doubles. This
shows that the pressure of a fixed mass of a gas increases with a decrease in volume at a constant
temperature. This can be demonstrated in Figure 6.3.
Figure 6.2 graph of pressure of the gas against volume
95
When the volume of the container containing the gas is reduced, pressure increases because
decreasing the volume of the container increases the impact on the walls of the container. Hence
force increases that cause an increase in pressure.
Boyle’s law states that the pressure of a fixed mass of gas is inversely proportional to its
volume provided the temperature of the gas is kept constant.
P α 1
V
PV = Constant, T
If pressure is plotted against 1/v, the graph is a straight line and it passes through the origin, as
shown in Figure 6.4:
At constant temperature
P1 x V1 = P2 x V2
Pressure
1
Volume
Figure 6.4 pressure is directly proportional to 1/volume
Figure 6.3 relationship between volume and pressure
96
Worked example
At a constant temperature, the pressure of a gas with volume 20 m
3
is 2000 Pa. Calculate the
volume of the gas if its pressure is 5000 Pa.
Solution
P1 = 2000 Pa P2 = 5000 Pa
V1 = 20 m
3
V2=?
P1 x V1 = P2 x V2
2000 x 20 = 5000 x V2
2000 x 20 = V2
500
V2 = 80 m
3
97
Charles’ Law
Charles was relating volume and temperature of a gas at constant pressure. Charles’ law was
found in 1787 by Jacques Charles.
RESULTS
When the water in the container is heated, the length of the trapped air column also increases.
The results can be plotted on a graph as shown in Figure 6.6.
Experiment 6.2
AIM: To investigate the relationship between volume and temperature at constant pressure.
MATERIALS: Thermometer, container, rubber band, oil, water, capillary tube and heat source
PROCEDURE:
1. Set up the experiment as shown in Figure 6.5 below.
2. Heat the water in the container.
3. Check the change in the length of the trapped air column.
Figure 6.5
98
EXPLANATION
From the graph in Figure 6.6, it shows that:
It is a straightline graph.
Volume is halved when temperature is halved.
Volume is doubled when temperature is doubled.
Volume of a fixed mass of a gas increases with an increase in temperature at a constant pressure
because the kinetic energy of the molecules increases. This makes the molecules move further
apart. When the molecules move further apart the volume occupied by the molecules increases.
Charles’ Law states that the volume of a fixed mass of gas is directly proportional to its
absolute temperature provided the pressure of the gas is kept constant.
V α T
V = constant, which is P.
T
At constant pressure,
V1 = V2
T1 T2
Volume
| | |
0 273 373 K
-273 0 100
0
C
Temperature (K/
0
C)
Figure 6.6 graph of volume against temperature
99
Worked example
At constant pressure, the temperature of 40 cm
3
of a gas is 55
0
C. What is temperature of 80 cm
3
of the gas?
Solution
V1 = 40 cm
3
V2 = 80 cm
3
T1 = 55+273= 328 K T2 =?
V1 = V2
T1 T2
40 = 80
328 T2
40 x T2 = 80x 328
T2 = 80 x 328
40
T2 = 656 K or 383
0
C
100
Pressure law
Pressure law relates pressure to temperature at constant volume. Pressure law which is also
known as Gay-Lussac’s law was found by Joseph Louis Gay-Lussac in1809.
Experiment 6.3
AIM: To investigate the relationship between pressure and temperature at constant volume.
MATERIALS: Container, sources of heat, thermometer, flask, water, tubing, Bourdon
pressure gauge, thermometer.
PROCEDURE:
1. Set up an experiment as shown in Figure 6.7.
2. Heat the water in the container.
3. Record the temperature and pressure readings.
Figure 6.7
101
RESULTS:
When water is heated, both temperature and pressure increases. The results can be used to plot a
graph as shown in Figure 6.8 below:
EXPLANATION
Increasing the temperature of a gas increases the kinetic energy of gas molecules. Hence the
force at which the molecules bombard the sides of the container increases since P=F/A.
From the graph in Figure 6.8, it shows that:
It is a straightline graph.
Pressure is halved when temperature is halved.
Pressure is doubled when temperature is doubled.
Pressure law states that the pressure of a fixed mass of a gas is directly proportional to its
temperature at constant volume.
P α T
P = constant, which is V
T
P1 = P2
T1 T2
Figure 6.8 graph of pressure of the gas against temperature
102
Whereby:
P1is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Worked example
At a constant volume, the pressure of a gas is 1.5 x 10
5
Pa and temperature is 80
0
C. What will be
the new pressure if the temperature has decreased to 25
0
C?
Solution
P1 = 1.5x10
5
Pa P2 =?
T1 = 80 +273= 353K T2 = 25 + 273 =298K
P1 = P2
T1 T2
1.5 x 10
5
= P2
353 298
298 x 1.5 x 10
5
= P2
353
P2 = 126629 Pa or 1.3 x 10
5
Pa
The combined gas equation
The combined gas law or general gas law is an equation formed by the combination of the three
gas laws, and shows the relationship between the pressure, volume and temperature for a fixed
mass of gas.
Gas equations can be summarized as follows:
Pressure law: P = constant, which is V.
T
Boyle’s law: PV = constant, which is T.
Charles’ law: V = constant, which is P.
T
103
For a fixed mass of gas, the combined gas equation becomes;
PV = constant
T
Initial gas law is given as:
P1V1 = constant
T1
After changes the equation changes into;
P2V2 = constant
T2
These two equations are related as follows;
P1V1 = P2V2
T1 T2
Whereby: P1 is pressure before change in any appropriate unit provided the same unit is
used on both sides of the equation.
V1 is volume before change in any appropriate unit provided the same unit is used
on both sides of the equation.
T1 is temperature before change in Kelvin (K).
P2 is pressure after change in any appropriate unit provided the same unit is used
on both sides of the equation.
V2 is volume after change in any appropriate unit provided the same unit is used
on both sides of the equation.
T2 is temperature after change in Kelvin, K.
Worked Examples
1. A cylinder has a volume of 0.12 m
3
and contains nitrogen gas at a pressure of 1620 Pa and
temperature of 20
o
C. After some of the gas has been consumed, it is found that the pressure has
fallen to 1100 pa and the temperature is then 10
o
C. Determine the volume of the gas.
Solution
V
1
=0.12m
3
P
1
=1620 Pa T
1
= 20+173K = 193K
104
V2=? P2=1100 Pa T2= 10+173=183K
P1V1 ═ P2V2
T1 T2
1620 Pa x 0.12m
3
═ 110Pa x V2
193K 183K
1.007 = 0.601 xV2
V2= 1.007
0.601
V2 =1.68 m
3
2. A tyre has a volume of 0.055 m
3
and contains air at a pressure of 145 Pa and a temperature of
280K. What temperature does the tyre have if its pressure is to be increased to 196 Pa assuming
the volume of the tyre has increased to 0.52 m
3
?
V1 = 0.055 m
3
T1=280 K P1 = 145 Pa
V2 = 0.52 m
3
T2=? P2 =196 Pa
Solution
P1 xV1 = P2 x V2
T1 T2
145 Pa x 0.055 m
3
= 196 Pa x 0.52 m
3
280 K T2
0.028 = 101.92
T2
0.028 x T2 = 101.92
T2 = 101.92
0.028
T2 = 3640K
105
6.2 Applications of gas laws
Gas laws can be applied in the following:
1. Bicycle pump
A bicycle pump uses Boyle’s law, which states that the pressure of the fixed mass of a gas
increases with a decrease in volume at constant temperature.
2. Car tyre
When a car is travelling at a high speed its tyres get inflated. This happens because the gases
inside the tyre get heated and collide on the walls of the tyre with greater force and, most often,
which causes more pressure to be exerted on the walls of the tyre.
3. Scuba diving
Scuba stands for self-contained underwater breathing apparatus. Divers carry tanks of
compressed gas to breathe under water. As they dive deeper, the water exerts pressure on their
bodies and the tanks. The air in the tanks has to be regulated and the pressure reduced so that it is
the same as the pressure of the surrounding water. The volume of air in their bodies decreases as
the pressure increases, using Boyle’s law. This makes the divers to descend quickly.
Exercise
In your groups, answer the following questions:
1. Explain the following:
a. Pressure of a fixed mass of a gas increases with a decrease in volume at a constant temperature.
b. Volume of a fixed mass of a gas decreases with a decrease in temperature at a constant pressure.
2. A gas syringe contains 80 cm
3
of oxygen gas at 50
0
C. If the temperature was increased to 80
0
C,
calculate the new volume of the gas, assuming constant pressure throughout.
3. A gas syringe contains 50 cm
3
of air at a pressure of 3 atmospheres. If the pressure was decreased to
1.5 atmospheres, calculate the new volume of the gas, assuming constant temperature throughout.
4. The volume of a gas at 40
0
C and pressure 1x10
5
Pa was 60 m
3
. Calculate the volume of a gas at 0
0
C
and pressure 3x10
5
Pa.
106
4. The constant volume gas thermometer
The constant volume gas thermometer is similar to the apparatus used to establish the pressure
law. It uses pressure law, which states that the pressure of a gas increases with an increase in
temperature at a constant volume, in order to measure the temperature.
6.3 Measuring gas pressure
Manometer is used to find gas pressure.
The manometer shown in Figure 6.9 is used to measure gas pressure.
The levels of mercury in columns A and B are the same because they experience the same
standard atmospheric pressure (760 mm Hg).
Figure 6.9 the level of mercury before opening the gas supply.
Figure 6.10 shows the same manometer after opening the gas supply.
107
The level of mercury in A has decreased while in B it has increased because gas pressure has
pushed mercury downwards in A and forced mercury upwards in B.
For the mercury to rise in B, at first it was equal to atmospheric pressure, then it overcame the
atmospheric pressure.
Gas pressure equals atmospheric pressure plus difference of levels of mercury in A and B.
Difference in the levels of mercury = h
Gas pressure = Atmospheric pressure + h
Gas pressure = 760 mm Hg + h
A manometer is used to measure lung pressure by blowing in gas from your lungs.
Lung pressure = 760 mm Hg + h
Worked example
Figure 6.11 is a diagram of a manometer used to measure gas pressure. The readings are in cm
Hg.
a. Read the pressure difference in mmHg.
b. Calculate the pressure of the gas supply if the atmospheric pressure is 760 mm Hg.
Solution
a. h = 6 cm Hg = 60 mm Hg
b. Pressure of the gas supply = standard atmospheric pressure + pressure
difference
= 760 mm Hg + 60 mmHg
Pressure of the gas supply = 820 mm Hg
Figure 6.11
108
Summary
The properties of the fixed mass of an ideal gas are affected by temperature, pressure and
volume.
Boyle’s law states that the pressure of a fixed mass of a gas is inversely proportional to volume
when temperature is kept constant.
P1V1 = P2V2
Charles’ law states that volume of a fixed mass of a gas is directly proportional to temperature
when pressure is kept constant.
V1 V2
T1 T2
P1 P2
T1 T2
The general equation for the general gas law is given as:
P1V1 = P2V2
T1 T2
Gas laws are used in bicycle pump, car tyre, scuba diving and constant volume gas thermometer.
A manometer is an instrument which is used to measure lung pressure or air pressure.
Student assessment
1. State the
a. Boyle’s law
b. Charles’ law
c. Pressure law
2. Explain, using the kinetic theory, the following:
a. Pressure of a fixed mass of a gas is inversely proportional to volume at constant
temperature.
b. Volume of a fixed mass of a gas is directly proportional to temperature at constant
pressure.
=
=
109
c. Pressure of a fixed mass of a gas is directly proportional to temperature at constant
volume.
3. Table 6.1 shows results obtained from an experiment to verify pressure law.
Temperature (
0
C)
25
30
35
40
45
Pressure (Pa)
10
20
30
40
50
Table 6.1
a. Plot a graph to show the relationship between pressure and volume.
b. How can you tell that the graph obeys Boyle’s law?
c. State two variables that were kept constant in this investigation.
d. From the graph, what is the temperature at the pressure of 32.5 Pa? Show on the
graph how you have found the answer.
4. A student sets up a mercury barometer at the top of a mountain. She finds that the length
of the mercury column is 0.70 m. Calculate the atmospheric pressure in Pa. (density of
mercury = 13600 kg/m
3
and g = 10 m/s
2
).
5. Figure 6.12 shows a sealed glass syringe that contains air and many very tiny suspended
dust particles.
Figure 6.12
a. Explain why the dust particles are suspended in the air and do not settle to the bottom.
b. The air in the syringe is at a pressure of 1.0 × 10
5
Pa. The piston is slowly moved into
the syringe, keeping the temperature constant, until the volume of the air is reduced from
55 cm
3
to 25 cm
3
. Calculate the final pressure of the air.
6. Hydrogen gas in a container has a volume of 0.11 m
3
at a pressure of 1200 Pa and
temperature of 15
0
C. What will be the new volume of the gas at 10
o
C and pressure of 970
Pa?
110
7. A balloon, volume 0.4 m
3
, containing oxygen gas at a pressure of 1x10
5
Pa is released
from the ground when the temperature is 17
o
C.Volume of a gas changes to 0.53 m
3
and
pressure becomes 0.5x10
5
Pa. Calculate its new temperature.
8. Describe an experiment that you would carry out to investigate the relationship between
volume and temperature of a fixed mass of dry air at constant pressure.
9. A measuring mass of a gas is placed in a cylinder at atmospheric pressure as shown in
Figure 6.13.
Figure 6.13
Explain what happens to volume and pressure when the gas is compressed. In each case
give a reason to your answer.
10. Driving a car raises the temperature of the tyres. What effect will this have on the
pressure of the air in the tyres? Give a reason for your answer.
11. 2 l of a gas exerts 5 atm of pressure. Calculate the pressure exerted by the same gas at
constant temperature when its volume is 5 l.
12. Explain two applications of gas laws.
13. A gas syringe contains 70 cm
3
of oxygen gas at 25
0
C. If the temperature was increased to
50
0
C, calculate the new volume of the gas, assuming constant pressure throughout.
14. A gas syringe contains 100 cm
3
of air at a pressure of 6 atmospheres. If the pressure was
decreased to 3 atmospheres, calculate the new volume of the gas, assuming constant
temperature throughout.
15. The volume of a gas at 30
0
C and pressure 1x10
5
Pa was 50 m
3
. Calculate the volume at
0
0
C and pressure 2.5x10
5
Pa.
111
16. With the aid of well labeled diagrams, explain how a manometer could be used to
measure lung pressure.
17. Figure 6.14 is a diagram of an instrument used to measure gas pressure.
a. Name the instrument.
b. Read the pressure difference in mmHg.
c. Calculate the pressure of the gas supply if the atmospheric pressure is 750 mm Hg.
Figure 6.14
cm Hg
112
Scalar and Vector quantities
Objectives
At the end of chapter 7, you must be able to:
Define scalar and vector quantities
Represent vectors
Add and subtract vectors
Resolve vectors
7.1 Scalar and vector quantities
Physical quantities can be analysed by dividing them into scalar quantities and vector quantities.
Scalar quantities
Scalar quantities are quantities that give the magnitude (size or numerical value) only.
For example, these could be distance, mass, length, height and temperature.
Addition and subtraction of scalar quantities
Scalar quantities are added or subtracted algebraically since they have no effect on direction.
Worked examples
1. 10 metres of cloth plus 5 metres of cloth
Solution
Total length of cloth = 10 m + 5 m
Total length of cloth = 15 m
2. 100 kg of salt minus 50 kg of salt
Chapter 7
113
Solution
Total mass of salt = 100 kg – 50 kg
Total mass = 50 kg
Vector quantities
Vector quantities are quantities that have both magnitude and direction.
For example, force, displacement, velocity and acceleration.
Distance and displacement
Why is distance a scalar quantity while displacement is a vector quantity?
Distance and displacement are two quantities that may seem to mean the same thing yet have
distinctly different definitions and meanings.
Distance refers to how much ground an object has covered during the motion.
For example: 10 m. Therefore, distance is a scalar quantity because it has magnitude (size) only.
Exercise 7.1
In your groups, answer the following questions:
1. Categorise each quantity as being either a vector quantity or a scalar quantity.
10 km, 10 m/s, 15 minutes, 100
0
C, 256 bytes, 80 N, 18 years old, 10 m towards north,
10 km/h eastwards and 10 m/s
2
.
2. Complete the table below to identify the physical quantities as scalars or vectors.
Physical quantity
Scalar or vector
Area
Speed
Acceleration
Velocity
Temperature
Kinetic energy
Force
114
Displacement refers to how far and out of place an object is. It is an object’s overall change in
position. For example: 10 m eastwards. Therefore, displacement is a vector quantity because it
has both magnitude and direction.
Worked example
Cecilia walked 10 m due west, 4 m due north, 10 m due east then 4 m due south as shown in
Figure7.1.
Calculate
a. the distance she covered.
b. the displacement during her journey.
Solutions
a. the distance she covered = 10 m + 4 m + 10 m + 4 m = 28 m.
b. Total displacement = 4 m + 10 m – 4 m – 10 m – 0 m
Total displacement = 0 m
(Displacement is 0 m because 10 m west is cancelled with 10 m east and 4 m north is
cancelled with 4 m south. She will go back to the same starting point. Therefore, there is
no displacement).
7.2 Representation of a vector quantity
The magnitude of a vector quantity is represented by a straight line while the direction is
represented by an arrow.
For example, a force of 10N can be represented as
10 m
4 m 4 m
Finishing point
Starting point 10 m
Figure 7.1
115
The vector quantity can also be represented as follows:
From Figure 7.2 above:
The magnitude of a vector is 10 m/s
The direction of the vector is at an angle of 50
0
from north line.
7.3 Addition and subtraction of vectors
When adding or subtracting vectors, the final vector is called the Resultant Vector.
10 N
Sketch Tail Head
Scale diagram
Using a scale of 1 cm to represent 1N, we can draw a line of length 10 cm.
10 cm
N
10 m/s
50
0
Figure 7.2
Exercise 7.2
In your groups, draw the following vectors to scale:
1. 6 N 2. 4200 N 3. 100 N 4. 1200 N
5. Velocity of 100 m/s at an angle of 60
0
to the east of the north line.
116
Addition and subtraction of in-line vectors (e.g. forces)
If two forces act in the same direction, their combined effect or resultant force is obtained by
joining forces head to tail.
Worked example
To find the resultant force when a 10 N and a 5 N forces are acting in the same direction.
Solution
10 N 5 N
+
Joining the forces head to tail:
10 N 5 N
The resultant force becomes:
15 N
Therefore, the resultant force = 15 N to the right
If forces are in opposite directions, the resultant force is obtained by subtracting the smaller force
from the larger force.
Worked example
To find the resultant force when the forces stated above act in opposite directions.
Solution
10N
5 N
The resultant force = 10 N – 5 N
The resultant force = 5 N in the direction of 10 N force (to the right).
Vectors at an angle to each other
Vectors can act at an angle to each other. When vectors are acting at an angle to each other, the
resultant vector is the final vector. The resultant vector can be found by using either triangle rule
or parallel rule.
117
1. Triangle of forces rule
Triangle method of resolving forces is made by considering forces that act at the same point and
on the same plane.
Worked examples
1. Find the resultant of two forces when 5N and 10 N act at right angle to each other.
Solution
Join the forces head to tail
The resultant is the line that completes the triangle and the direction is represented by angle ѳ
towards the direction of a greater force (10 N).
Calculation
Using Pythagoras’ theorem
R
2
= 10
2
+ 5
2
R
2
= 100 + 25
5 N
10 N
Figure 7.3
R 5N
ѳ
10 N
Figure 7.4
118
R
2
= 125
R = √125
R = 11.18N
The resultant force is 11.18 N to the 10 N force at an angle ѳ.
To find angle ѳ, use tangent
Tan ѳ = opp = 5 N
Adj 10 N
Tan ѳ = 0.5
Angle ѳ = tan
-1
0.5
Angle ѳ = 26.6
0
The Resultant force is 11.18N at an angle of 26.6
0
to the horizontal ground in the direction of a
10 N force.
Drawing a scale diagram
Using a scale of 1cm = 1N
10 N = 10 cm, 5 N = 5 cm
Measuring R with a ruler.
R = 11.2 cm
Resultant force = 11.2 x 1N
Resultant force = 11.2 N
Measure angle ѳ using a protractor
Angle ѳ = 26.5
0
The Resultant force is 11.2 N at an angle of 26.5
0
to the horizontal ground in the direction of
a 10 N force.
2. Two forces of 800 N and 600 N act on an object at an angle of 70
0
, find the resultant force.
R
5N (5 cm)
ѳ
10 N (10 cm)
Figure 7.5
119
Solution
Join the forces head to tail
Calculation
To find R using cosine Rule
R
2
= 600
2
+ 800
2
– 2 x 600 x 800 x cos110
0
R
2
= 360 000+ 640 000 – (-328 339.3376)
R
2
=1 000 000 + 328 339.3376
R
2
= 1 328 339. 338
R = √1 328 339.338
R = 1 152.5 N
Resultant force is 1 152.5 N to the 800 N force at an angle ѳ
To find angle ѳ
Using sine Rule
sin ѳ = sin 110
0
600 1 152.5
600N
70
0
800 N
Figure 7.6
R 600 N
ѳ 110
0
70
0
800 N
Figure 7.7
120
sin ѳ = 600 X sin110
0
1 152.5
sin ѳ = 0.4892
Angle ѳ = sin
-1
0.4892
Angle ѳ = 29.3
0
Resultant force is 1 152.5N at 29.3
0
to the horizontal ground in the direction of 800 N force.
Drawing a scale diagram
Using a scale of 1cm = 100 N
600 N = 6cm
800 N = 8cm
Measure the length of R using a ruler
R = 11.53 cm
Resultant force = 11.53 X 100 N
Resultant force = 1153 N
Measure angle ѳ using a protractor
Angle ѳ = 29.2
0
Therefore, Resultant force is 1153 N at an angle of 29.2
0
to the horizontal ground towards 800 N
force.
R 600 N (6 cm)
ѳ 70
0
800 N (8 cm)
Figure 7.8
121
2. Parallelogram method
Parallelogram method is used to find the resultant of two forces acting at a given point.
In a parallelogram rule, the forces complete the parallelogram. The resultant force is represented
by a diagonal of the parallelogram. The angle between the diagonal and a horizontal force gives
the direction of the resultant force.
Worked examples
1. Two forces of 6 N and 8 N are acting at an angle of 30
0
. Find the resultant force by
calculation and drawing a scale diagram.
Solution
Draw in two more lines to complete the parallelogram. Then draw a diagonal line, and calculate
its length.
The diagonal represents the resultant of the two forces.
6 N
30
0
8 N
Figure 7.8
6 N
R
30
0
ѳ 150
0
30
0
8 N
Figure 7.9
122
Calculation
To find R, using cosine rule
R
2
= 6
2
+ 8
2
– 2 x 6 x 8 x cos150
0
R
2
= 36 + 64 – (-83.1384)
R
2
= 100 + 83.1384
R
2
= 183.1384
R = 183.1384
R = 13.5N
Resultant force is 13.5N to the 8N force at an angle ѳ.
To find ѳ, using sine rule
Sin ѳ = sin 150
0
6 13.5
sin ѳ = sin150
0
x 6
13.5
sin ѳ = 0.2222
Angle ѳ = 12.8
0
Resultant force is 13.5 N at 12.8
0
to the horizontal ground in the direction of 8 N force.
Drawing a scale diagram
Using a scale of 1 cm = 1 N
6 N = 6 cm
8 N = 8 cm
6 N (6cm) R
30
0
ѳ 30
0
8 N (8 cm)
Figure 7.10
123
Measure the length of R using a ruler
R = 13.5 cm
Resultant force = 13.5 X 1 N
Resultant force = 13.5 N
Measure angle ѳ using a protractor
Angle ѳ = 12.8
0
Therefore, Resultant force is 13.5 N at an angle of 12.8
0
to the horizontal ground towards 8N
force.
3. James walks 5 km due north then 7 km due east. Draw a scale diagram to find the
resultant of his displacement.
Solution
Scale: 1 cm to represent 1 km
5 km = 5 cm
7 km = 7 cm
The length of R = 8.6 cm
Resultant displacement = 8.6 X 1km
Resultant displacement = 8.6 km
Angle ѳ = 54.5
0
Therefore, resultant displacement is 8.6 km at an angle of 54.5
0
from the north line.
7 km (7 cm)
5 km (5 cm)
R
ѳ
Figure 7.11
124
Resolving components of vectors
Previously, two forces acting at a point were used to find a single force called resultant force. In
this section we are going to look into the reverse of that process. In reversing the process, a
single force called resultant force can be replaced by two forces having the same effect.
Therefore, a force is said to be resolved. A force can be resolved into two components namely:
vertical component
horizontal component
These components of forces are perpendicular (at a 90
0
angle) to each other.
The vector sum of these two components is equal to the original force in magnitude and
direction. These two forces must also pass through the same point of application as the original
force.
Exercise 7.3
In your groups, draw scale diagrams to find the resultant force in each case using
1. Triangle technique
2. Parallelogram technique
a. b.
4 N 3500 N
50
0
7200 N
6N
c.
600 N
100
0
800 N
Figure 7.12
125
To calculate the values of forces acting at right angles as shown in Figure 7.13:
The vertical component = Resultant force x sine Ө
FV = F Sin Ө
The horizontal component = Resultant force x cosine Ө
FH = F Cos Ө
Worked example
Khataza pulls Tegha sitting in a trolley by using a string. The tension of the string is
100 N inclined at 60
o
to the horizontal. Calculate:
a. the horizontal force pulling Tegha in the trolley.
b. the vertical force tending to lift the trolley.
Solution
Since we know that tension is100 N. This force is the resultant F. The sketch can be drawn as
shown in Figure 7.14.
a. horizontal force FH = F cos Ө
FH = 100 N x cos 60
o
Vertical
component (FV)
Resultant
force (F)
ѳ
Horizontal component (FH)
Figure 7.13 components at right angle.
FV
F = 100 N
60
0
FH
Figure 7.14
126
FH = 50 N
b. Vertical force FV = F sin Ө
FV = 100 N x sin60
o
FV = 86.6 N
Summary
A scalar quantity is a quantity which has magnitude (size) only.
A vector quantity is a quantity which has magnitude (size) and direction.
Distance is a scalar quantity because it has magnitude only while displacement is a vector
quantity because it has both magnitude and direction.
Vector quantities are represented by vector diagrams.
The resultant vector is the vector which is found after adding or subtracting vectors.
In-line vectors are added when acting in the same direction and subtracted when acting in
opposite directions.
The resultant vector for vectors that are acting at an angle to each other is found using either
triangle rule or parallelogram rule.
Student assessment
1. Define the following:
a. Resultant vector
b. Component of a force
Exercise7. 4
In your groups, answer the following questions:
1. Draw a scale diagram of a 50 N force at 60
0
to the horizontal and show
that its vertical and horizontal components are 43.3 N and 25.0 N
respectively.
2. Resolve a 20 N force acting at an angle α = 30
0
to the horizontal into
a. vertical component
b. horizontal component
127
2. Explain the difference between
a. Scalar quantity and vector quantity.
b. Distance and displacement.
3. Explain why speed is a scalar quantity while velocity is a vector quantity.
4. Show if the following are scalar or vector quantities:
Quantity
Scalar or vector
Temperature
Deceleration
Density
Velocity
Mass
Speed
Table 7.1
5. James travels 200 m north, then turns and travels 80 m east.
a. Draw a scale diagram representing this trip.
b. Determine what his actual displacement is.
6. A 15 N force and a 6 N force act in line. Draw scale diagrams to find the resultant force if
a. they act in the same direction.
b. they act in opposite directions.
7. A 3 N force and a 5 N force act at an angle of 60
0
to each other. Calculate the resultant
force.
8. Figure 7.15 shows two forces acting at a right angle.
Draw a scale diagram to determine the resultant force, by using:
a. triangle rule.
3 N
5 N
Figure 7.15
128
b. parallelogram rule.
9. Calculate the resultant force produced by forces of 7 N and 3 N acting on a point object,
if the lines of action of the forces are:
a. at a right angle to one another.
b. at 80
0
to one another.
10. A 20N and a 50N force both act at the same point.
a. If two forces act at a right angle to each other, find by scale drawing the size and
direction of their resultant force. (Scale; 1cm: 5N).
b. If two forces act at 60
o
to each other, draw a scale diagram using parallelogram technique
to find the size and direction of the resultant force.
11. Linda walks 50 m north, 20 m east, 50 m south then 20 m east.
Calculate:
a. the distance she covered.
b. her displacement.
12. Draw a scale diagram showing a displacement of 360 N at 50
0
from the north line
eastwards.
13. Tikondane pushes a box as shown in Figure 7.16.
a. By calculation, find:
i. the vertical component
ii. the horizontal component.
b. By drawing a scale diagram, find
i. the vertical component of a force
ii. the horizontal component of a force.
80 N
45
0
Figure 7.16
129
14. Figure 7.17 shows a 30 N force acting to the horizontal.
a. Complete the vertical and horizontal components.
b. Calculate:
i. the vertical component of a force.
ii. the horizontal component of a force.
Figure 7.17
130
Linear motion
Objectives
At the end of chapter 8, you must be able to:
Describe distance, displacement, speed, velocity
and acceleration
Conduct experiments to determine velocity and
acceleration
Determine acceleration due to gravity
Explain motion-time graphs
Apply equations of uniformly accelerated motion
8.1 Distance, displacement, speed, velocity and
acceleration
Distance and displacement
Distance
Distance refers to how much ground an object has covered during the motion. Distance is a
scalar quantity because it has magnitude only.
For example: 10 km.
Displacement
Displacement is the distance moved in a particular direction. Displacement is a vector quantity
because it has both magnitude and direction.
For example: It is a displacement of 20 km due west.
Chapter 8
131
Speed
A car can go fast or slow. When a car goes fast it means its speed is high and when it goes slow
it means its speed is low. In each case, to describe speed we use distance covered and time taken
to cover that distance.
So, speed is defined as the distance covered per unit time. You can also define speed as the rate
of change in distance.
Speed = Distance covered
Time taken
A speed of 1m/s is realised when a distance of 1m is covered in 1 second.
S = D
T
Whereby D is distance in metres (m), T is time in seconds (s) and S is speed in metres per
second (m/s).
Therefore, the SI unit for speed is m/s.
If distance D is in kilometers (km) and time T is in hours (hrs), then speed S is given in
kilometers/hour (km/hr).
Speed is a scalar quantity since it has magnitude only.
Worked example
An athlete covered a distance of 20 km in 5hours. Calculate the speed of the athlete.
Solution
To find the speed of the athlete, use the covered distance and time taken for that distance to be
covered.
Distance = 20 km, t = 4 hours
Speed = distance moved
Time taken
Speed = 20km
4hrs
Speed = 5 km/hr
D
S T
132
Average Speed
When an athlete was running at different speeds or his speed was varying, his average speed can
be worked out as follows;
Average speed = Total distance covered
Total time taken
Worked examples
1. A cyclist covers the first 90 km of the distance traveling at 30 km/hr then he covers the next
80 km traveling at 40 km/hr. Calculate the average speed of the cyclist.
Solution
The first part of the journey
D = 90 km S = 30 km/hr
T= D
S
T = 90 km
30 km/hr
Experiment 8.1
AIM: To find the speed of a moving object
MATERIALS: Chalk, stop watch, trolley tape measure
PROCEDURE:
1. Find a smooth and flat area around your school.
2. Measure a distance of 5 m with a tape measure then mark the starting and
finishing points with chalk.
3. One observer must be on the starting line with a stop watch and a trolley
while the other observer must be on the finishing line.
4. The observer on the starting line must push the trolley and start the
stopwatch immediately.
5. As the trolley crosses the finishing line, a second observer must raise
his/her hand so stopwatch is stopped immediately.
6. Use the formula to find the speed of a trolley:
Speed = distance (5 m)
Time taken
133
T = 3hrs
Second part of the journey
D = 80km S = 40km/hr
T = 80km
40km/hr
T = 2hrs
Average speed = Total distance covered
Total time taken
Average speed = 90 km + 80 km
3 hrs + 2 hrs
Average speed = 170 km
5 hrs
Average speed = 34 km/hr
2. A car starts at a speed of 0 m/s until it reaches a speed of 10 m/s. Find the average speed of the
car.
In this case the average speed = sum of the speeds
Number of speeds
= 0 m/s + 10 m/s
2
= 10 m/s
2
Average speed = 5 m/s
134
Velocity
Velocity is the distance covered in a stated direction (displacement) per unit time.
Therefore, velocity is the speed in stated direction. Velocity, like speed, is measured in m/s.
Velocity is the vector quantity because it has magnitude and direction of travel.
Velocity is the distance covered in a stated direction per unit time or displacement per unit time.
Velocity = Distance covered in a stated direction (displacement)
Time taken
Worked example
A car travels 300 m in 20 s. What is its velocity?
Solution
Velocity = Distance covered in a stated direction
Time taken
Velocity = 300 m
20 s
Velocity = 15 m/s
Experiment 8.2
AIM: To find the average speed of an athlete
MATERIALS: Tape measure, stop watch and whistle.
PROCEDURE:
1. Measure a distance of 20 m with a tape measure and mark the starting and finishing
points.
2. An athlete must stand on the starting point.
3. An observer must have a stop watch and a whistle.
4. The athlete must start running as soon as the observer blows the whistle and starts the
stop watch.
5. The athlete can cover the same distance three times then the observer stops the stop
watch.
6. Record the total distance covered as 20 m x 3 = 60 m.
7. Record the total time taken.
8. Calculate the average speed of the1 athlete as follows:
Average speed = Total distance (60 m)
Total time taken
135
Acceleration
An object accelerates when its velocity changes.
Acceleration is the rate of change of velocity per unit time. Acceleration can also be defined as
the change in velocity per given time.
Acceleration = Change in velocity
Time taken
Let the initial velocity be u
The final velocity be v
Time to be t
Change in velocity becomes = final velocity (v) – initial velocity (u)
The acceleration becomes:
a = v – u
t
When v is greater than u, an object speeds up. Speeding up is called acceleration.
When v is less than u, an object slows down. Slowing down is called deceleration.
When v is equal to u, an object has zero acceleration (travels with a constant velocity) because
there is no change in velocity.
Acceleration, like velocity is a vector quantity.
Worked examples
1. A motor cycle starts from rest (0 m/s) and reaches a velocity of 10 m/s in 5 seconds.
Calculate the acceleration of the motor cycle.
Solution
u = 0 m/s v= 10 m/s t = 5 s
a = v – u
t
a = 10 m/s – 0 m/s
5 s
a = 0 m/s
5 s
a = 2 m/s/s or 2 m/s
2
or 2 ms
-2
Acceleration of a motorcycle is 2 m/s
2
136
2. An aeroplane wants to land. Its velocity drops from 20 m/s until it reached 5 m/s in
10 seconds. Calculate the average acceleration of the aeroplane.
Solution
u = 20 m/s v = 5 m/s t = 10 s
a = v – u
t
a = 5 m/s – 20 m/s
10 s
a = -15 m/s
10s
a = -1.5 m/s
2
NOTE the negative sign means that the acceleration is in the opposite direction to the chosen
direction of the velocity. The negative acceleration is called Deceleration or Retardation.
Therefore, in the above example,
Deceleration = - acceleration
Therefore, Deceleration = 1.5 m/s
2
137
Experiment 8.3
AIM: To determine velocity and acceleration of an object
MATERIALS: chalk, stopwatch, trolley, tape measure
PROCEDURE:
1. Find a smooth and flat area around your school.
2. Measure distances of 2 m, 5 m and 9 m with a tape measure then mark these points
including the starting point on 0 m and finishing point on 9 m with chalk.
3. Four observers must be on the starting line (0 m), 2 m, 5 m and 9 m with stop watches.
4. The observer on the starting line must push the trolley and start the stopwatch
immediately.
5. Each observer must start the stopwatch as the trolley crosses each point.
6. Use your results to complete the Table 8.1.
Section (m)
Time at start(s)
Time
interval(s)
Length of
section (m)
Velocity (m/s)
0-2
2
2-5
3
5-9
4
Table 8.1
7. Use the results in Table 8.1 to calculate the velocity of a trolley
using a formula
velocity = displacement (length of a section)
time interval
8. Plot a graph of velocity against time
9. Calculate the acceleration of the trolley for each section by using the gradient of the
velocity time graph.
Gradient = change in velocity
Change in time
138
Determing the velocity and acceleration of an object by using
a tickertape-timer
The tickertape-timer marks dots on a tape at regular intervals of 1/50 s(0.02s). This is taken from
the frequency of alternating mains electricity because a ticker-time uses alternating current. In
this case the frequency is 50 Hz (50 cycles per second).
Figure 8.1 shows a tickertape-time used to investigate the motion of a trolley.
When a trolley is connected to the tape and set in motion, the dots will be created on the tape.
The pattern of dots acts as a record of the trolley’s movement. The time interval between
adjacent dots is 1/50 s (0.02 s).
Even spacing on the tape: constant velocity
Increasing spacing: increasing velocity
The distance from the start to the fifth dot is covered at an interval of 1/10 s (0.1 s). This will be
considered as the section which represents the trolley’s displacement.
Measure and record the distance (displacement) of every fifth dot from the start of the tape.
The velocity of a trolley for the first section can be calculated by using a formula:
Velocity = displacement (length of the section)
Time interval (0.10 s)
Figure 8.1 tickertape timer
139
Repeat the measurement and calculation for the other two sections.
Now we can record the values in the Table 8.2 for the three sections.
Section of a
tape
Time at start (s)
Time interval(s)
Displacement(cm)
Velocity (m/s)
1
0.0
0.1
4.0
0.40
2
0.1
0.1
8.0
0.80
3
0.2
0.1
12.0
1.2
Table 8.2 An example of the graph obtained using a tickertape-timer
NOTE: The displacements must be converted to metres first.
Use the results in the Table 8.2 to plot a velocity-time graph.
We can calculate the acceleration of a trolley by calculating the gradient of the graph.
Acceleration = gradient of a graph = change in velocity
change in time
Velocity(m/s)
Change in velocity
Change in time
Time(s)
Figure 8.2 graph of velocity against time
140
8.2 Acceleration due to gravity
All the objects that are near the earth surface fall freely. They do not experience any air
resistance. They fall under the force of gravity and they fall with uniform acceleration. This
acceleration is called the acceleration due to gravity. The acceleration due to gravity is also
called the acceleration of free fall.
Free fall is the falling of an object with uniform acceleration under the force of gravity if air
resistance is negligible.
Acceleration due to gravity is denoted a letter g. The value of acceleration due to gravity is
approximately 10 m/s
2
. This value varies slightly from one place to another on the earth’s
surface. This is because the gravitational pull of the earth on an object also varies. The variation
of acceleration due to gravity is less than 1%.
Exercise 8.1
Individually, answer the following questions:
1. Explain why speed is a scalar quantity and velocity is a vector quantity.
2. A body moves a distance of 10 m in 5 s. Calculate the speed of the body in
a. m/s
b. km/h
3. A bicycle rider accelerates from rest to a velocity of 30 m/s in 10 s. Calculate the acceleration of
the rider.
4. An athlete changes her speeds uniformly from 30 m/s to 20 m/s in 5 s. Calculate her retardation.
5. a. Plot a velocity time graph using the values in Table 8.2.
b. Calculate the acceleration of the trolley
141
RESULTS/EXPLANATION
From the graph of T
2
against l, a straight line through the origin should be obtained.
The gradient of the line gives a value of g.
Experiment 8.4
AIM: To measure acceleration due to gravity, g
MATERIALS: Stop watch, bob, metre ruler, clamp stand, retort stand and string.
PROCEDURE:
1. Set up an experiment as shown in Figure 8.3.
2. Pull the bob 5 cm to one side and release it so that it swings in one plane.
3. As the bob passes from left to right across your centre line begin to count 3, 2, 1, 0.
Start the stop watch as you count 0.
4. Record the time taken for 50 oscillations.
5. Repeat the experiment for the length 80, 60, 40 and 20 cm long.
6. In each case the timing must be repeated as a check on the previous reading and the
results can be recorded in Table 8.2.
Length l
(cm)
Time for 50 oscillations
Periodic
time T (s)
T
2
T
2
l
1
2
mean
100
80
60
40
20
Table 8.2
7. Plot a graph of T
2
(y-axis) against l (x-axis) and measure its slope.
Support
String (100 cm)
Bob
Figure 8.3
142
The value of g can also be calculated from the results obtained as follows:
T
2
4π
2
l g
Therefore, g = 4π
2
÷ T
2
l
8.3 Graphical representation of motion
It is very difficult to be precise in describing the motion of the objects in motion. Therefore,
graphs are used in science to assist our understanding and description of objects in motion.
Graphs are a convenient and accurate means of displaying information.
Graphs can be plotted as follows:
displacement (distance) against time
velocity (speed) against time
Distance–Time Graph
In a distance–time graph, distance or displacement (in the y–axis) is plotted against time (in the
x–axis).
The gradient of a distance-time graph = change in y (distance)
change in x (time)
Distance = speed
Time
Therefore, the gradient of a distance–time graph gives speed.
If the graph is plotted, displacement against time, its gradient gives velocity.
Distance
Change in y
Change in x
Time
Figure 8.4 distance-time graphs.
=
143
Interpreting the distance (displacement)–time graphs
The motion of the object under the distance –time graph can be described as follows:
Worked example
The graph in Figure 8.6 below represents the distance covered by Wadada express bus during
the first 6 hours of its motion. Use it to answer the questions that follow.
1. Describe the motion of the bus at parts A, B and C.
d d
Uniform or constant speed (velocity) Stationary (no speed)
t t
d Increasing speed (velocity) d
Decreasing speed (velocity)
t t
Figure 8.5 distance (displacement) – time graphs
Distance (km)
20
15 C
B
10
5 A
0 0 1 2 3 4 5 6
Time (hrs)
Figure 8.6
144
2. Determine the speed of the bus during the first 2 hours.
3. Calculate the average speed of the bus.
Solutions
1. Motion of the bus:
Part A: the bus is travelling at uniform or constant speed
Part B: the bus is stationary
Part C: the bus is travelling at uniform or constant speed
2. Speed = Distance covered
Time taken
Speed = 10 km
2 hrs
Speed = 5 km/hr
3. Average speed = Total distance
Total time taken
= 20 km
6 hrs
Average speed = 3.3 km/hr
Velocity (speed)–time graph
In velocity–time graph velocity (in the y–axis) is plotted against time (in the x–axis).
Velocity
(m/s)
V
Change in velocity
Time
U
Time (s)
Figure 8.7 Velocity–time graph
145
Gradient = change in velocity
Time
Change in velocity = acceleration
Time
Therefore, a velocity–time graph gives acceleration.
Acceleration from the graph can also be found by using a formula:
a = V– U
t
Describing the motion on a velocity – time graph
v v
uniform (constant) acceleration Uniform (constant) deceleration
t t
v v
Increasing acceleration
Zero acceleration
Uniform (constant) velocity or speed
t t
v
Decreasing acceleration
t
Figure 8.8 velocity- time graphs
146
Distance under Velocity (Speed)–Time Graph
Distance =Area under velocity–time graph
Worked examples
1. Use a speed-time in Figure 8.9 to find the total distance traveled by an object from A to
D.
Solution
Total distance =Area of ∆ X + Area of rectangle Y + Area of ∆ Z
= (½ x b x h) + (l x w) + (½ b x h)
= (½ x 2 s x 40 m/s) + (40 m/s x 2 s) + (½ x 2 s x 40 m/s)
= 40 m + 80 m + 40 m
The total distance covered by an object = 160m.
Total distance can also be worked out as follows:
Total distance = Area of a trapezium
= ½ (a + b) h
= ½ (2 + 6) x 40
= ½ x 8 x 40
Total distance = 160 m
Speed m/s
40 B C
30-
20-
10- X Y Z
0A | | | | | |D
0 1 2 3 4 5 6
Time (s)
Figure 8.9
147
2. Figure 8.10 is a speed – time graph for a cyclist. Use it to answer the questions that
follow.
a. Describe the motion of the cyclist during the entire 10 seconds.
b. Calculate the acceleration of the cyclist during the first 4 seconds.
c. Calculate the total distance covered during the entire 10 seconds.
d. Calculate the average speed of a cyclist during the entire 10 seconds.
Solution
a. The motion of a cyclist
0 – 4 s = the cyclist accelerates uniformly
4 – 10 s = the cyclist had zero acceleration or the cyclist had a uniform speed
b. Acceleration during the first 4 seconds
a = Change in speed
time taken
a = V – U
t
U = 0 m/s
Figure 8.10
148
V = 8 m/s
t = 4 s
a = 8 m/s – 0 m/s
2s
= 8 m/s
4 s
Acceleration = 2 m/s
2
c. Total distance = Area of a trapezium.
= ½ (a + b) h
= ½ (6 + 10) x 8
= ½ x 16 x 8
Total distance = 64 m
d. Average speed = total distance
total time
= 64 m
10 s
Average speed = 6.4 m/s
3. A body starts from rest and accelerates at 10 m/s
2
for 5 s. It then continues at this speed for 5 s
before decelerating to rest in 10 s.
a. Sketch a speed time graph of this motion.
b. Calculate the distance the object moved in the first 5 seconds.
Solution
a. Sketch of a speed time graph
149
b. Distance moved in 5 seconds = Area of a triangle
= ½ x b x h
= ½ x 5 x 10
Distance = 25 m
8.4 Motions of falling bodies
Three forces acting on a falling body are:
Gravitational force or weight (W) – It acts downwards
Speed (m/s)
10
0
0 5 5 10
Time (s)
Figure 8.11
Exercise 8.2
Individual work:
A trolley is pushed uniformly for 5 seconds from a velocity of 0 m per second to a
velocity of 40 m per second, it continues at this steady velocity of 40 m per second
for a further 30 seconds and then decelerates uniformly for 10 seconds so that it
stops.
1. Draw to scale a velocity–time graph to represent the motion of the trolley.
(Scale1cm = 5 m/s and 1cm = 5 s).
2. From your graph, calculate:
a. Total time taken for the journey.
b. Acceleration during the first 5 seconds,
c. Distance covered during the first 35 seconds
d. Deceleration during the last 10 seconds.
d. Deceleration during the last 10 seconds
150
Upthrust (U) – It is the push by the fluid. It acts upwards.
Frictional force (Fr) – It opposes the motion. It acts upwards.
Acceleration of a free fall
In air, a coin falls faster than a feather because they experience different size of air resistance.
Air resistance is greater to lighter bodies than to heavy ones. In a vacuum a coin and a feather
fall at the same rate because they do not experience any air resistance. The coin and a feather are
said to have a free fall.
U
Fr
W
Figure 8.12 forces acting on a falling object
Object
151
RESULT/EXPLANATION
When the feather and a coin are dropped from the same height and at the same time, the coin
reaches the bottom of the tube first because it experiences less air resistance since it is heavier
than the feather.
Experiment 8.5
AIM: To investigate free fall of bodies
MATERIALS: Feather, coin, glass tube, vacuum pump and cork.
PROCEDURE:
1. Set up the experiment as shown in Figure 8.13 below.
2. Drop the feather and a coin in a glass tube. Observe which one will reach the bottom of the
tube first.
Give a reason for this result.
3. Connect a vacuum pump and pump out all the air.
4. Drop the feather and a coin from the same height. Observe which one will reach the bottom
of the tube first.
Give a reason for this result.
Discuss your results with your friends in class.
Glass tube
To a vacuum pump
Figure 8.13
152
Vacuum pump connected:
When the feather and a coin are dropped from the same height at the same time, both the coin
and the feather will reach the bottom of the tube at the same time because they do not experience
any air resistance. They are falling under a free fall.
Figure 8.14 shows the falling of a coin and a feather in a vacuum and air.
Falling of heavy objects near the earth
In the 16
th
century the Italian scientist Galileo dropped a small iron ball and a large ball ten times
heavier from the top of the Learning Tower of Pisa. In this story, we were told that, to the
surprise of onlookers who expected the cannon ball to arrive first, both objects reached the
ground almost at the same time.
From this story, untrue we now think, suggests that the heavy bodies, whatever their sizes, are
only slightly affected by air resistance. Therefore, heavy objects near the earth fall under free-
fall.
Figure 8.14 falling in a vacuum and air
153
Falling in parachutes
Falling when a parachute is not opened
A parachutist falls at a very high speed, because W – (U + Fr) is very great since Fr is taken as
negligible.
As a parachutist increases the speed or accelerates frictional force increases until W = (Fr + U).
When W = Fr + U, a parachutist falls at a constant or uniform speed called terminal speed or
terminal velocity of 50 m/s.
This terminal velocity without opening a parachute is called Sky Diving.
Falling when a parachute is opened
After opening a parachute, frictional force increases which reduces the speed of a parachutist.
A parachutist decelerates since Fr has increased.
(Fr + U) > W
As the speed decreases, it causes Fr to decrease until (Fr + U) = W.
When (Fr + U) = W, a parachutist travels at another terminal speed or terminal velocity of 8 m/s.
This is a landing speed. Hence a parachutist lands safely on the ground.
Figure 8.15 Tower of Pisa
154
8.5 Equations of uniformly accelerated motion
Calculations involving the displacement, velocity, acceleration and time of motion of a moving
body use the equations of motions. These equations are derived from the definitions of
acceleration and average velocity.
Equation 1
If a body is moving with uniform acceleration a and its velocity increases from u to v in time t,
the equation is given as;
a = v – u
t
Making v the subject of the formula, the equation that is obtained is
v = u + at ………………………… (1)
Equation 2
The velocity of a body moving with uniform acceleration increases steadily. Its average velocity
therefore equals half the sum of its initial and final velocities. The equation is given as:
average velocity = u + v
2
Figure 8.16 graph of a parachutist.
155
From (1) v = u + at
average velocity = u + u + at = 2u + at
2 2
= u + ½ at
2
If s is the distance moved in time, t, then since average velocity = distance / time = s/t
s = u + ½ at
t
s = ut + ½ at
2
…………………………. (2)
Equation 3
The third equation is obtained by eliminating t between the first two equations.
Squaring both sides of the equation, v = u + at, we obtain
v
2
= u
2
+ 2 uat + a
2
t
2
Taking out the factor 2a from the last two terms of the right-hand side,
v
2
= u
2
+ 2a (ut + ½ at
2
)
But the bracket term is equal to s
Hence v
2
= u
2
+ 2as ………………………. (3)
Worked examples
1. Yusuf rides a bicycle. He starts from rest and accelerates at 2 m/s
2
for 10 seconds.
Calculate his maximum speed.
Solution
u = 0m/s v =? a = 2m/s
2
t = 10 s
v = u + at
v = 0 + (2 x 10)
v = 20 m/s
2. Fatima throws an apple vertically upwards with an initial velocity of 15 m/s. Neglecting
air resistance, and taking the acceleration due to gravity as 10 m/s
2
, calculate:
a. the maximum height reached by the apple
b. the time taken before the fruit reaches the ground.
Solution
a. u = 15 m/s v = 0 m/s a = -10 m/s
2
(when an apple is thrown upwards it
decelerates)
s =?
156
v
2
= u
2
+ 2as
2as = v
2
– u
2
s = v
2
– u
2
2a
s = 0
2
– 15
2
2 x (-10)
s = -225
-20
s = 11.25 m.
b. s = 11.25 m u = 0 m/s a = 10 m/s
2
t =?
s = ut + ½ at
2
11.25 = (0 x t) + (½ x 10 x t
2
)
11.25 = 5 t
2
t
2
= 11.25
5
t
2
= 2.25
t = √2.25
t = 1.5 s
Summary
Distance and speed are scalar quantities because they have magnitude only while displacement
and velocity are vector quantities because they have magnitude and direction.
Exercise 8.3
Individually, answer the following questions:
1. A motorist starts from rest and accelerated uniformly at the rate of 5 m/s
2
for 5 seconds.
Calculate
a. the final speed reached
b. the distance covered.
2. A fruit falls from rest from the tree. Ignoring air resistance and take acceleration due to
gravity = 10 m/s
2
, calculate
a. the velocity after 4 s.
b. the distance covered after 2 seconds.
157
Speed is defined as distance covered per given time. Velocity is the distance covered in a stated
direction (displacement) per given time.
Acceleration is the rate of increase in velocity while deceleration is the rate of decrease in
velocity.
a = v – u
t
d = -a
All the objects that are near the earth surface fall freely and they fall under the force of gravity
called acceleration due to gravity or acceleration of free fall.
Free fall is the falling of an object with uniform acceleration under the force of gravity if air
resistance is negligible.
Gradient of a distance–time graph gives speed or the gradient of a displacement–time graph
gives velocity.
Gradient of a speed–time graph or velocity–time graph gives acceleration.
The total distance under a speed (velocity)–time graph is found by calculating the area under the
graph.
The equations of uniformly accelerated motion are:
v = u + at …………(1)
s = ut + ½ at
2
……….(2)
v
2
= u
2
+ 2as…………(3)
Student assessment
1. Define the following:
a. Speed
b. Acceleration
c. Acceleration due to gravity
d. Free fall
2. Explain why speed is a scale quantity while velocity is a vector quantity.
3. Kelson walks from Naotcha to Chimwankhunda, 1.5 km distance in 30 minutes. Find his
average speed in
a. km/h
158
b. m/s
4. A car starts from rest and accelerates to a speed of 24 m/s in 6 seconds. Calculate its
acceleration.
5. A body changes its speed from 80 m/s to 40 m/s in 10 s. Calculate its retardation.
6. A bicycle rider accelerates from rest to a velocity of 30 m/s in10 s. Calculate the
acceleration of the rider.
7. If a body moves a distance of 10 m in 5s, then calculate the speed of the body.
8. Elita is running at a speed of 10 m/s speeds up uniformly to a speed of 20 m/s in 5 s.
Calculate her acceleration.
9. Figure 8.17 shows a graph of speed against time of an object thrown vertically upwards
and after 3 seconds returns to the same position downwards.
a. Describe the movement of an object in the first 3 seconds.
b. From the graph, what time is the speed of an object
i. greatest
ii. least
c. Calculate the acceleration of the object as it falls back towards the ground.
d. Calculate the greatest distance above the ground reached by the drop.
Speed (m/s)
40 -
30 –
20 –
10 –
0 | | | | | |
0 1 2 3 4 5 6
Time (s)
Figure 8.17
159
10. Figure 8.18 shows a distance–time graph of an object.
a. Describe the motion of an object during the first 2 hours.
b. What is the total distance travelled by the object?
c. What is the total time taken by the object to cover the distance?
d. Calculate the speed of the object during the first 1 hour.
e. Calculate the average speed of the object during the entire journey.
11. a. Plot the appropriate distance-time graph from the results given.
Distance (m)
0
10
20
30
40
Time (s)
0
2
4
6
8
Table 8.4
b. From your graph, calculate the speed during 8 seconds.
12. A car, initially at rest, moves with uniform acceleration for 10 seconds until it attains a
velocity of 30 m/s. It then proceeds at this velocity for 20 seconds and finally comes to
rest after retarding uniformly for a further 5 seconds.
a. Draw a velocity–time graph of the motion
b. From the graph calculate
i. The deceleration of a car during the last 5 seconds
ii. The total distance moved by the car.
Distance (km)
40-
30
20
10
0- | | | |
08:00 09:00 10:00 11:00 12:00
Time of the day
Figure 8.18
160
13. Figure 8.19 is a velocity–time graph of Chitundu Luxury Coach. Use it to answer the
questions that follow.
a. Describe the motion of the coach during the entire 14 minutes.
b. Calculate the acceleration during the first 2 minutes.
c. Calculate the retardation during the last 4 minutes.
d. Calculate the total distance covered.
14. State the difference between;
a. Scalar quantity and vector quantity.
b. Distance and displacement
c. Speed and velocity.
15. A car travelling at 5 m/s is uniformly accelerated at 8 m/s
2
for 7 seconds. Calculate the
distance covered by the car.
16. A particle is sliding down a slope with a uniform acceleration of 5 m/s
2
. If its initial
velocity was 3 m/s, calculate its velocity after it has slid 20 m down the slope.
17. A ball is dropped vertically downwards at an initial speed of 10 m/s. Calculate its
velocity after 10 s. (Assume g = 10 m/s
2
).
Figure 8.19
161
Work and energy
Objectives
At the end of chapter 9, you must be able to:
Calculate work done
Explain the conservation of mechanical energy
Solve problems related to work and energy
9.1 Work
Work is done when a force produces motion. In physics work is defined if force applied on
object displaces the object in the direction of force. The greater the force and the greater the
distance moved, the more work is done.
For example: an example is when you are running, when you carry a load up a ladder and when
a car is moving.
Work is said to be done when a force moves its point of application in the direction of the force.
Work done = force x distance moved by force in the direction of the force
If force is measured in Newton (N), distance in metres (m), then work is measured in Nm or
Joule (J).
1 Joule of work is done when a force of 1 N moves an object 1 m in the direction of the force.
Chapter 9
1 N force
moves
distance
1 m
Figure 9.1 demonstrating work done
Object
Object
162
Worked examples
1. Ramazan provides a force of 50 N to move an object a distance of 100 cm. Calculate the
work done.
Solution
F = 50 N d = 100 cm = 1 m
W = F x d
W = 50 N x 1 m
W = 50 Nm or 50 J
2. Yankho has a mass of 80 kg. Calculate the work done by Yankho in climbing a ladder
5 m high.
Solution
Force = weight = mg
Force = 80 kg x 10 m/s
2
= 800 N
d = 5 m
W = F x d
= 800 N x 5 m
W = 4000 J
Exercise 9.1
In your groups, answer the following questions:
1. Find the work done in each case. Show your working:
a. A 20 N force moves an object 10 m.
b. A 50 kg bag of flour is lifted 2.5 m.
c. A 200 g mass moves a distance of 50 cm.
2. The work done by a 10 N force to move an object is 30 J. Calculate the distance
covered by an object.
163
Work done by a force acting at an angle
A force, F, can act on a body so as to move it in a direction other than its own. This situation can
occur only if there is some other force preventing motion taking place in the direction of a force.
An example is a man pulling a garden roller as shown in Figure 9.2.
In Figure 9.2, the man is holding the handle at an angle ѳ to the horizontal and exerts a force F
in the direction shown. The work done by the force F in the direction of motion is found by using
a formula:
W = F x cos ѳ
Experiment 9.1
AIM: To find the amount of work done
MATERIALS: Ruler or tape measure, spring balance, scale, masses, a wall, a ladder
or stairs and a bench.
PROCEDURE:
1. Tie a rope to the mass and suspend it to the spring balance to find its weight or
force in Newton (N). Use a ruler or tape measure to measure the distance from
the ground to the top of a bench in metres (m). Lift the mass from the ground
to the top of the bench. Calculate the work done (w = f x d).
2. Step on a scale to find out your mass, then convert your mass from kilograms
to Newtons by simply multiplying by 10. Use a ruler or a tape measure to
measure the length of a ladder. Lean the ladder against a wall and climb it up
to the end. Calculate the work done (w = f x d).
F
ѳ
Direction of motion
Figure 9.2 a man pulling a roller
164
Worked example
A lady applies a force of 60 N to move a vacuum cleaner at an angle of 60
0
to the horizontal.
Calculate the work done.
Solution
F = 60 N Angle ѳ = 60
0
W = F x cos ѳ
= 60 N x cos 60
0
W = 30 J
9.2 Conversion of mechanical energy
Energy can be converted from one form to another. A common conversion of mechanical energy
is from potential energy to kinetic energy or vice versa. When there is conversion of these
mechanical energies, some energy is usually wasted in form of heat or sound.
This can be demonstrated by using a pendulum.
When a bob is made to oscillate, it converts kinetic energy to potential energy by moving to
points A and C. The potential energy is maximum at points A and C. Kinetic energy is maximum
at point B. This means that the energy of the bob is all potential energy at A and C and it is all
kinetic energy at B.
In this case, Kinetic energy and potential energy are interchangeable continually.
The energy changes can be summarised as follows:
A to B to C: Potential energy to kinetic energy to potential energy
A C
B
Figure 9.3 a pendulum
165
After some time, the bob fails to reach positions A and C, because potential energy changes to
heat energy due to friction between the bob and the air particles.
Law of conservation of mechanical energy
From the pendulum, it is noticed that the loss in PE of a pendulum equals the gain of the KE and
vice versa.
When a bob oscillates from point A to point B, its potential energy at point A equals its kinetic
energy at point B. When the bob oscillates from point B to point A, its kinetic energy at point B
equals its potential energy at point A. The total mechanical energy is kept constant during this
oscillation. This means there is neither increase nor decrease in mechanical energy. Energy is
not lost or created; it simply changes from one form to another. Therefore, mechanical energy is
conserved and it is summarized as the law of conservation of energy.
Law of conservation of energy states that energy is neither created nor destroyed, but it can
simply change from one form to another.
Energy–work theorem
Work is done whenever a force moves an object.
Work done = force x distance moved.
Energy is the ability to do work. Things have energy in order to do work. Whenever work is
done, energy is transformed.
For example, if you lift a 20kg box to a height of 5 m, the work done by lifting the box will be:
W = F x d
W = 200 N x 5 m
W = 2000 J
In this case, the box will gain a potential energy of 2000 J, assuming there is no air resistance.
If this box is dropped to the ground, 2000 J of work is done in accelerating the box.
The box losses 2000 J of potential energy. If the box is about to hit the ground, 2000 J of kinetic
energy is gained. If the box hits the ground and comes to rest, 2000 J of kinetic energy is
changed into heat energy.
Therefore, work done equals energy
Work done = Energy
Energy is measured in Joules (J).
166
Worked example
A 1.5 kg brick is lifted from the ground to a height of 3 m.
Calculate
a. the work done in lifting the brick.
b. the energy used in lifting the brick.
c. the potential energy gained by the brick after being lifted to a height of 3 m.
Solution
a. W = F x d
W = 150 N x 3 m
W = 450 J
b. E = W
E = 450 J
c. PE = W
PE = 450 J
Summary
Work is said to be done when a force moves its point of application in the direction of the force.
Work done by a force acting in the direction of motion is found by using a formula:
Work done = force x distance moved by force in the direction of the force
SI unit for work done is Joule (J).
Work done by a force acting at an angle to the horizontal direction of the motion is found by
using a formula:
Work done = force x cosine ѳ
Law of conservation of mechanical energy states that energy is neither created nor destroyed but
it simply changes from one form to another.
SI unit of energy is Joule (J)
Energy- work theory is given as:
Work = Energy
167
Student assessment
1. Define
a. Work
b. Energy
2. Explain the similarity between work done and energy.
3. State the law of conservation of mechanical energy.
4. Describe energy changes that take place in a vibrating spring.
5. Figure 9.4 is a diagram of a simple pendulum. The mass vibrates between the points X
and Z through Y.
a. At which point is:
i. Potential energy maximum?
ii. Potential energy minimum?
iii. Kinetic energy minimum?
iv. Kinetic energy maximum?
b. What happens to the potential energy when mass oscillates from:
i. X to Y?
ii. Y to X?
iii. Z to Y?
iv. Y to Z?
c. What happens to the kinetic energy when mass oscillates from:
i. X to Y?
ii. Y to X?
iii. Z to Y?
X Z
Y
Figure 9.4
168
iv. Y to Z?
d. Describe the energy changes that take place when a mass oscillates from
i. X to Y
ii. Y to X
iii. Z to Y
iv. Y to Z
v. X to Y to Z
e. Explain why a mass would eventually stop oscillating.
6. Calculate the work done when a 10kg bag is lifted to a height of 2 m.
7. How much work is done if a force of 80 N moves an object a distance of 5 m?
8. A 50 N force is used to lift an object. Calculate the distance covered by the object if the work
done on it is 100 J.
9. What is the total work done when 10 bricks of mass 1.5 kg each are lifted to a height of 3 m?
10. Christopher used 20 J of energy to lift a book from the ground up on to a shelf.
a. Calculate the potential energy of the book when it is on the shelf.
b. State the work done to lift the book up on to the shelf.
c. When the book falls from the shelf, how much kinetic energy does it have just before it hits
the ground? (Assume the air resistance is negligible).
d. What happens to this form of energy when the book hits the ground?
11. Carol raised a 40kg mass to a height of 2 m above the ground.
a. Calculate the potential energy of the mass after being raised to a height of 4 m.
b. What is the work done by Carol on the mass?
c. If the mass falls, calculate the kinetic energy:
i. when it is half-way down.
ii. just before impact with the ground.
169
12. Figure 9.5 is a diagram of a mass hanging on a spring. If the mass is pulled to point R and
released, it vibrates between points R and P.
a. At which point does the mass have
i. highest kinetic energy?
ii. lowest kinetic energy?
b. Explain the energy changes that take place when the mass is vibrating from P to R.
Figure 9.5
170
Machines
Objectives
At the end of chapter 10, you must be able to:
Describe what machines are
Explain efficiency, mechanical advantage and velocity
ratio of a machine
Calculate efficiency, mechanical advantage and
velocity ratio of machines
10.1 What are machines?
The term ‘machine’, makes many people think it is a complicated piece of mechanism. The term
‘machine’ has tended to lose its original meaning. It does not matter how a machine is deemed
complicated, but there are a limited number of basic mechanical principles:
In a machine forces are involved in energy conversions. Therefore, a machine is a device
that causes a change in the way that these forces act.
Machines can help to raise heavy objects with a smaller effort. Therefore, a machine is a
device that changes the magnitude of a force and makes work to be done easier.
In a machine the direction of a force changes; therefore, a machine is a device that
changes the direction that a force acts on.
From the principles explained above, machines are considered to change either the magnitude or
direction of a force.
In physics, a machine is any device in which a force applied at one point can be used to
overcome a force at some other point.
Examples of machines are levers, pulleys and inclined planes.
Chapter 10
171
The lever
A lever is any rigid body which is pivoted about a point called the fulcrum. Examples of levers
are claw hammer, wheelbarrow, pliers, nut crackers, sugar tongs, table knife and scissors.
In levers, a force called the effort is applied at one end to overcome a force called the load at
the other end. Levers use the principle of moments as discussed in book 2, chapter 6 and section
6.6.
A lever is used as a force multiplier because it uses a smaller effort to move a larger load. In
ancient times, humans used levers to lift very heavy objects like stones. It was believed that a
person can move the earth with a lever.
Pulleys
A pulley is a grooved rim (rims) mounted in a framework called a block. The effort is applied to
a rope, chain or belt which passes over the pulleys.
Figure 10. 1 levers
Pliers
Tongs
Scissors
Wheelbarrow
Crowbar
172
Inclined plane
An inclined plane is a plane surface at an angle to the horizontal. It is easier to move a heavy
object up an inclined plane than to move it vertically upwards.
Figure 9.2 pulley systems
Figure 10.2 pulley systems
Figure 10.3 inclined plane
Exercise 10.1
In your groups, use the following words to complete the statements:
A-inclined plane B-lever C-effort D-load E-simple machine
1. A simple machine that has a flat, slanted surface is an __________.
2. A simple machine that has a bar pivoting around a fulcrum is called a ________,
3. The object being moved by a lever is _______________. A ____________ has
only a few parts. The ___________________is the force used to do work.
Fixed pulley
Movable pulley
Fixed and movable pulley
173
10.2 Efficiency, mechanical advantage and velocity
ratio of a machine
Efficiency of a machine
Efficiency of a machine is the ratio of the work done by the machine to the total work put into
the machine expressed in percentage.
Work done by the machine on the load = load x distance load moves. This work done is called
work output.
Work put into the machine by the effort = effort x distance effort moves. This work done is
called work input.
Efficiency = work output (load x distance load moves) x 100 %
work input (effort x distance effort moves)
In a perfect machine, efficiency is 100 %. A perfect machine is a theoretical machine, with a
useless load of zero. Useless load is the force needed to overcome the frictional forces between
the moving parts of a machine to raise any of its moving parts. The efficiency of 100 % means
the work output equals work input.
Using a perfect machine, if an effort of 100 N is moved at a distance of 2m to raise a 200 N
force:
Work put into the machine by effort = work input = 100 N x 2 m = 200 J
Work done by the machine on the load = work output = 200 J
The load will move a distance of 1m
In practical machines, efficiency is always less than 100 %. Some work is always wasted to
overcome the frictional force between the moving parts of a machine and raise any of its moving
parts. The efficiency is less than 100 % because the useful work done by the machine is less than
the work put into the machine by the effort.
Mechanical advantage of the machine
The Mechanical advantage of a machine is the ratio of the two forces, the load and the effort.
The Mechanical advantage of a machine is found by dividing load by effort.
Mechanical advantage =
Load
Effort
174
Where mechanical advantage is greater than 1, it means the machine is designed to overcome a
load which is greater than the effort. An example is a car jack used to lift a motor car.
When mechanical advantage is less than 1, it means the machine is designed so that the effort
used is greater than the load. For example, a bicycle has a mechanical advantage of less than 1.
This can be noticed when the cyclist is cycling uphill where more effort is applied to work
against the force of gravity. The cyclist is said to be working at a mechanical disadvantage. The
cyclist simply dismounts and walks.
Velocity ratio of a machine
The velocity ratio of a machine is the ratio of the distance moved by the effort to the distance
moved by the load in the same time. It has no units.
Velocity ratio =
In a situation where mechanical advantage is greater than 1, velocity ratio is greater than 1
because the effort moves through a much greater distance than the load.
The velocity ratio of a machine can also be called speed ratio.
Relationship between mechanical advantage, velocity ratio and frequency:
work = force x distance
efficiency = x 100 %
But = Mechanical advantage (M.A.)
= x 100 %
Therefore, efficiency = M.A. x x 100 %
Efficiency = x 100 %
distance moved by the effort
distance moved by the load in the same time
Load x distance the load moves
Effort x distance the effort moves
Load
Effort
distance the load moves
distance the effort moves
1
velocity ratio (V.R.)
1
velocity ratio (V.R.)
M.A
V.R.
175
Worked examples
1. Figure 10.4 shows a wheelbarrow used to lift a load of 400 N.
Calculate
a. the mechanical advantage of the machine
b. the velocity ratio of the machine
c. efficiency of the machine.
Solution
a. MA =
=
MA = 4
b. VR =
=
VR = 5
c. Efficiency = x 100 %
= x 100 %
Figure 10.4
100 N
400 N
1m
0.2m
Load
Effort
400 N
100 N
distance moved by effort
distance moved by load
1 m
0.2 m
work output
work input
Load x distance moved by load
Effort x distance moved by load
176
= x 100 %
Efficiency = 80 %
OR
Efficiency = x 100%
= x 100%
Efficiency = 80%
2. Figure 10.5 shows a pulley system used to raise a 180kg mass. The load moves a distance of
5m.
Calculate:
a. the mechanical advantage, MA of the pulley system
b. the effort used to raise the load
c. the velocity ratio, VR for this system
d. the distance moved by the effort
e. work done by the effort
f. work done on the load.
400 N x 0.2 m
100 N x 1 m
MA
VR
4
5
Figure 10.5
177
Solutions
a. MA = number of pulley systems
MA = 4
b. MA =
Effort =
=
Effort = 450 N
c. VR = 4 (the number of ropes supporting the load)
d. VR =
Distance moved by effort = VR x distance moved by load
= 4 x 5 m
Distance moved by effort = 20 m
3. Figure 10.6 shows a 500 N object pushed to a lorry up the height 2 m using a plank of length
10 m.
Calculate
a. the mechanical advantage, MA of an inclined plane, if the effort is 200 N
Load
Effort
Load
MA
180 x 10 N
4
distance moved by effort
distance moved by load
Figure 10.6
2m
10m
178
b. the velocity ratio
c. the work done by the effort
d. the work done on the object.
Solution
a. MA =
=
MA = 2.5
b. VR =
=
VR = 5
c. work done by the effort = effort x distance moved by effort
= 200 N x 10 m
Work done by the effort = 2000 J
d. Work done on load = load x distance moved by load
= 500 N x 2 m
Work done on load = 1000 J
Load
Effort
500 N
200 N
distance moved by effort
vertical distance moved by load
10 m
2 m
179
Summary
Machines are devices that can reduce the effort required to move a load. Machines make work to
be done easier.
Examples of simple machines are inclined plane, levers and pulley.
Mechanical advantage of a machine =
In a pulley system:
MA = number of pulley systems
VR = the number of ropes supporting the load
Exercise 10.2
In your groups, answer the following questions:
1. Derive a formula which shows the relationship between efficiency, mechanical
advantage and velocity ratio.
2. Figure 10.7 shows 200 N load dragged up to the platform using an inclined plane and
using an effort of 40 N.
Calculate
a. the mechanical advantage, MA of the inclined plane.
b. the velocity ratio, VR of the plane.
c. the work done by the effort.
d. the work done on the load.
Load
Effort
Figure 10.7
200 N
3 m
6 m
40N
180
Velocity ratio =
Efficiency = work output (load x distance load moves) x 100 %
work input (effort x distance effort moves)
Efficiency can also be calculated from the formula:
Efficiency = x 100 %
Student assessment
1. Define the following:
a. Machine
b. Mechanical advantage
c. Velocity ratio
2. Derive a formula which relates efficiency, mechanical advantage and velocity ratio.
3. Figure 10.8 is a diagram of a lever.
a. Calculate the value of d.
b. Calculate the mechanical advantage of the machine, assuming there is no friction at
the fulcrum.
c. Calculate the velocity ratio of the lever.
d. Suggest one way of increasing the MA of the lever.
4. Explain why a lever is used as a force multiplier.
distance moved by the effort
distance moved by the load in the same time
MA
VR
Figure 10.8
100 N
10 N
d
10 m
181
5. Figure 10.9 is a diagram of a pulley system. The distance moved by the load is 10 m.
Calculate
a. the mechanical advantage of the pulley system.
b. the effort used to raise the load.
c. the velocity ratio.
d. the distance moved by the effort.
e. work done by the effort.
f. work done on the load.
6. Figure 10.10 is a diagram of an inclined plane.
Figure 10.9
Figure 10.10
200 N
E
182
a. Calculate the velocity ratio of the machine.
b. If the efficiency of the machine is 95%, calculate its mechanical advantage.
7. Figure 10.11 is a diagram of the simple two-pulley system used to lift 5 kg mass
a.
a. State the mechanical advantage of a pulley system.
b. State three ways of increasing the MA of the machine.
c. Calculate the distance the string at E must be pulled to lift the mass 0.5 m.
d. State the velocity ratio of the pulley system.
e. Calculate the efficiency of the pulley system.
f. What is the force at E required to lift the 5kg mass?
8. Catherine uses a rope to drag a box of weight 500 N up a smooth inclined wooden
plank of effective length 7 m and on to a platform 3.5 m high.
Calculate:
a. the effort she must exert on the rope
b. the velocity ratio
c. the mechanical advantage
d. the work done on the box in Joules.
9. Yamikani uses a rope to drag a box of mass 100 kg up an inclined plane of length 6 m
to a platform 1.5 m high.
a. Draw the arrangement of the inclined plane.
b. If he applies an effort of 200 N, calculate:
Figure 10.11
E
5 kg
183
i. the mechanical advantage of the machine
ii. the velocity ratio of the machine
iii. the efficiency of the machine.
c. How much work was done on the box as a result of being dragged up on the platform?
d. How much potential energy was gained by the box as a result of being raised to the
platform?
e. Calculate the amount of heat energy produced as a result of the friction between the
plank and the 100kg box.
f. State two ways of minimizing the amount of heat lost in this arrangement.
10. A pulley system has a velocity ratio of 2. Show where the load and effort are applied to
the pulley system.
a. Use the pulley system to complete Table 10.1.
Table 10.1
Load (N)
5
10
15
20
25
Effort (N)
1
2
3
4
5
MA
Efficiency (%)
b. Explain two reasons why the efficiency of the pulley system is less than 100 %.
11. Explain the characteristics that machines have in common.
184
Current electricity
Objectives
At the end of chapter 11, you must be able to:
Describe electric current
Describe potential difference
Describe electrical resistance
Analyse electric circuits
Determine electric power and energy
11.1 Electric current
The atoms in a solid are held together by strong electrical forces. These atoms can only vibrate
about a fixed mean position. A solid which is a conductor contains a great number of electrons
which are loosely held and are free to move. These are known as free electrons. When an
electric field is applied there is a drift of electrons in a conductor from the negative side to the
positive side. This movement of charge is known as electric current.
Therefore, electric current is the flow of electric charges (electrons) from the negative side of
an electric field to the positive side.
Figure 11.1 shows the direction of electrons or electric current in the circuit.
Chapter 11
+ - Direction of electric current
Figure 11.1 direction of electric current
185
The unit of electric current (I) is the Ampere (A). Therefore, the SI unit of current is Ampere
(A).
Current is measured by an ammeter in the circuit.
The symbol for an ammeter is:
Using an ammeter in the circuit to measure current
Connect the positive (red) terminal of an ammeter to a positive terminal of a power
supply (e.g. cell or battery) and the negative (black) terminal to a negative terminal of a
power supply. Any mistake on connection will break an ammeter.
Connect an ammeter in series circuit because it measures current passing through a
component or a wire. An ammeter has very low resistance and has a negligible effect to
the flow of current.
The smaller currents are measured by a milliammeter. The unit then is milliampere (mA).
1 A = 1000 mA
The quantity of electricity (electric charge) which passes any point in a circuit will depend on the
strength of the current and the time for which it flows. The quantity of electric charge is called
Coulomb.
A
Figure 11.2 an ammeter
186
A coulomb is the electric charge which passes any point in a circuit in 1 second when a steady
current of 1 ampere is flowing.
Using the following symbols:
Q for electric charge in Coulombs, C
I for current in Amperes, A
t for time in seconds, s
Electric current, I can be found as:
I =
From the above equation, we can define current as the rate at which the electric charge flows.
Worked examples
1. An electric charge of 50 C flows past a point in a wire in 5 seconds. Calculate the current
flowing in the wire.
Solution
Q = 50 C t = 5 s
I =
I =
I = 10 A
2. A current of 30 mA flows in a circuit for 100 seconds. Calculate the quantity of electric
charge.
Solution
I = 30 mA = 3 x10
-3
A t = 100 seconds
Q
t
Q
I
t
Q
t
50 C
5 s
187
Q = It
Q = 30 x 10
-3
A x 100 s
Q = 3 C
11.2 Potential difference (pd) or Voltage
When a cell is connected in the circuit, the electric charge flows because it has the energy called
potential or potential energy. In a circuit, potential is the energy associated with a charge at a
point in an electric field because of the force acting on it.
Work is done against the electric field when an electric charge is moved from a lower potential
level to a higher potential level. The difference between the two levels is known as the potential
difference (pd) or voltage.
Potential difference is defined as a difference in potential between two points, equal to the
energy change when a unit electric charge moves from one place to another in an electric field.
The SI unit of potential difference is the volt (V).
Exercise 11.1
In your groups, answer the following questions:
1. Convert 100 mA into Amperes (A)
2. How many milliamperes (mA) are in 0.003 A?
3. Calculate the electric charge delivered if a current of 20 A flows for
10 seconds.
4. Calculate the current in the circuit if a 120 C of electric charge
flows for 2 seconds.
188
Potential difference (which is also called voltage) is measured by a voltmeter.
The symbol for a voltmeter is:
Connecting a voltmeter
A voltmeter is connected in the following ways:
A positive side (red side) of a voltmeter is connected to the positive terminal of a cell or
battery or any power supply while a negative side (black side) is connected to a negative
terminal.
A voltmeter is connected in a parallel circuit, across a component that it is measuring its
voltage because it measures voltage between two points. A voltmeter has very high
resistance. Connecting a voltmeter in a series circuit prevents the flow of current.
Two points are at a potential difference of 1 volt if 1 Joule of work is done per Coulomb of
electric charge from one point to another.
Figure 11.3 a voltmeter
V
189
Potential difference is lost in the bulbs and none is lost in the connecting wire. The sum of the
potential differences lost in the bulbs equals the potential difference across the supply.
Sum of potential differences = 0.7 V + 0.3 V + 0.5 V = 1.5 V
Electromotive force (EMF)
Each cell or battery has the potential difference written across it, e.g. 1.5 V.
A cell or a battery produces its highest potential difference when it is not in the circuit and when
it is not supplying current. This maximum potential difference is called the electromotive force
(EMF) of the cell or a battery.
Electromotive force (EMF) is the maximum potential difference across a cell or battery when it
is not in a circuit and not supplying current (when I = 0 A).
Pd= 1.5 V
0.7 V 0.3 V 0.5 V
Figure 11.4 Measuring potential difference using a voltmeter
V
0.7 Joule of
energy is lost by
each Coulomb
0.3 J of
energy is lost
by each
Coulomb
0.5 Joule of energy
is lost by each
Coulomb
1.5 Joules of energy is given to
each Coulomb
Figure 11.5 measuring Emf
1.5 V
V
Emf = 1.5 V
190
When the cell is connected in the circuit and supplies current, the potential difference across the
cell drops because of energy wasted inside the cell. For example, a potential difference across a
1.5 V cell can be 1.3 V. Part of the energy is used to push electrons and overcome internal
resistance.
Exercise 11.2
In your groups, answer the following questions:
1. Explain the difference between potential difference and electromotive force.
2. Figure 11.6 is a circuit diagram showing voltmeters used to measure voltage
across the battery and the resistors.
a. Calculate the reading of the voltmeter V
3
.
b. How much energy is given to each Coulomb by the battery?
c. How much energy is lost by each Coulomb in:
i. R
1
? ii. R
2
?
Figure 11.6
2.5 V
R1
R2
191
11.3 Electrical resistance
In section 11.2, it was explained that current flows through the conductor or circuit because of
the potential difference (voltage which is applied across it.
Different materials have different conductivities when they are connected in the circuit. For
example:
In a copper wire current is high because electrons pass easily. This shows that a
copper wire is a good conductor. Therefore, copper has low resistance.
In a nichrome wire of a similar size as the copper wire gives low current because
electrons pass with difficulties. This shows that a nichrome wire is not a good
conductor. Therefore, nichrome wire has high resistance.
The conductivity of the wires mentioned above is different because the materials in the wires
provide different opposition to the flow of electrons. This opposition is called electrical
resistance.
Electrical resistance is the opposition to the flow of electrons in a wire or a circuit.
Therefore, copper wire has low electrical resistance while nichrome wire has high electrical
resistance.
Electrical resistance is measured in Ohms (Ω). Therefore, the SI unit of resistance is Ohm (Ω).
If the electric current through a conductor is I when the pd across it is V, its electrical resistance
can be calculated by the equation:
Electrical resistance =
R =
Worked example
The potential difference across a nichrome wire is 10 V. If the current flowing through the wire
is 2 A, calculate the electrical resistance of a nichrome wire.
Solution
Pd = 10 V I = 2 A
Potential difference
Current
V
I
192
R =
R = 10 V
2 A
R = 5 Ω
The ohm is the electrical resistance of a conductor in which the current is 1 A when a p.d. of 1 V
is applied across it.
Factors affecting the electrical resistance
There are four major factors that affect resistance of a wire.
These factors are length of a wire, temperature, cross-sectional area and nature of the material.
If you want to vary one factor, the other three factors must be kept constant.
V
I
193
Length of the wire
RESULTS
Resistance decreases as the length of the wire decreases from 100 cm going downwards.
EXPLANATION
As the length of the wire decreases, there are few collisions that take place between flowing
electrons and stationary positive ions. In other words, there is reduction in opposition to the flow
of electrons because there are few stationary positive ions that can cause resistance.
In general,
Shorter wire → lower resistance
Experiment 11.1
AIM: To investigate the effect of length of a wire on its resistance.
MATERIALS: Nichrome wire of length 100 cm, connecting wires, 2 cells, crocodile clips,
ammeter, metre ruler and voltmeter.
PROCEDURE:
1. Connect the circuit as shown in Figure 11.7.
Figure 11.7
2. Complete the circuit by connecting a 100cm nichrome wire across the gap AB.
3. Record the voltmeter and ammeter readings. Record your results in Table 11.1.
4. Repeat the experiment with other lengths, e.g. 80 cm, 60 cm, 40 cm and 20 cm. In each case
record the voltmeter and ammeter readings in the table.
5. Complete the values of resistance.
Length of
wire(cm)
Voltage(V)
Current(A)
Resistance(Ω)
100
80
60
40
20
Table 11.1
A B
A
V
194
Longer wire → higher resistance.
CONCLUSION
Therefore, the resistance of the wire varies directly with the length of the wire.
R α l
Figure 11.8 shows the shapes of the graphs that can be plotted from experiment 11.1.
Figure 11.8 graphs showing the effect of the length of the wire on resistance
Resistance
(Ω)
Length of the wire (cm)
(a) Resistance of the wire
increases with an increase
in length
Current
(A)
Length of the wire (cm)
(b) Current decreases with an
increase in length of the wire
195
Thickness or cross-sectional area
RESULTS
Resistance of the wires decreases as the thickness increases.
EXPLANATION/CONCLUSION
As the thickness of the wire increases, electrons are able to flow with less resistance.
Therefore, resistance of the wire varies inversely with its cross-section or thickness.
R α 1
A
Figure 11.10 shows the shapes of the graphs that can be plotted from experiment 11.9.
Experiment 11.2
AIM: To investigate the effect of thickness of the wire on its resistance.
MATERIALS: 4 Nichrome wires of the same length and material but with different cross-
sections or diameters, ammeter, voltmeter, connecting wires, crocodile clips and 2 cells.
PROCEDURES:
1. Set up the experiment as shown in Figure 11.9.
2. Complete the gap AB with the smallest wire. Figure 11.4
3. Record the voltage and ammeter readings.
4. Repeat the experiment with nichrome wire of different thickness. In each case record, the
voltmeter and ammeter readings.
Nichrome wire
Voltage(V)
Current (A)
Resistance(Ω)
1
2
3
4
Table 11.2
Calculate the resistance of a nichrome wire of each thickness by using the formula
R =
V
/
I
A B
Figure 11.9
A
V
196
Figure 11.10 graphs showing the effects of cross-sectional area of the wire on resistance
Temperature
Experiment 11.3
AIM: To investigate the effect of temperature of the wire on its resistance.
MATERIALS: Nichrome wire of length 10 cm, connecting wires, ammeter, voltmeter, 2
cells and crocodile clips.
PROCEDURE:
1. Set up an experiment as shown in Figure 11.11
Figure 11.11
2. Record the voltmeter and ammeter readings in the circuit.
Calculate the resistance of nichrome wire(R=
V
/I)
3. Light a Bunsen burner and place it at a distance of 20 cm from nichrome wire.
4. Record the new voltmeter and ammeter readings, and then calculate the new resistance of
the nichrome wire.
5. Repeat the experiment with the Bunsen burner being at distances, 15cm, 10cm and 5cm.
Record the voltmeter and ammeter readings then calculate the resistance in each case.
Distance from the
burner(cm)
Voltage(V)
Ammeter(A)
Resistance(Ω)
No burner
15
10
5
Table 11.3
Nichrome wire
A
V
R
(Ω)
A (cm
2
)
(a)
R
(Ω)
1 (1/cm
2
)
A
(b)
197
RESULTS
The resistance of the wire increases as the distance of the Bunsen burner from the nichrome wire
decreases.
EXPLANATION
As the distance between the burner and the wire decreases, the temperature of the wire increases.
An increase in the temperature of the wire increases the kinetic energy of the stationary positive
ions in the wire. These particles increase their vibrations and cause more collisions with the
flowing electrons. Hence increases resistance.
In general:
Low temperature → low resistance
High temperature → high resistance.
CONCLUSION
Therefore, resistance of the wire varies directly with temperature.
R α T
Figure 11.12 shows the shapes of the graphs that can be plotted from experiment 11.3.
Figure 11.12 Graphs showing the effect of temperature on resistance
Resistance
(Ω)
Temperature (
0
C)
(b) Resistance of the wire
increases with an increase
in temperature
Current
(A)
Temperature (
0
C)
(b) Current decreases with an
increase in temperature
198
Nature of the material
RESULTS
Copper wire gives high ammeter reading while nichrome wire gives low current reading.
EXPLANATION
Copper wire gives high current reading because it has low resistance. Nichrome wire gives low
current because it has high resistance.
This shows that copper wire is made up of copper material which has low resistance and
nichrome wire is made up of nichrome material which has high resistance.
CONCLUSION
Therefore, different wires are made up of different materials that have different resistances.
Experiment 11.4
AIM: To investigate the effect of the nature of the material on the resistance of
the wire.
MATERIALS: Copper wire and nichrome wire of the same thickness and
length, ammeter, voltmeter, 2 cells and crocodile clips.
PROCEDURE:
1. Set up an experiment as shown
in Figure 11.13.
Figure 11.13
2. Complete the gap with copper wire.
3. Record the voltmeter and ammeter readings.
4. Then calculate the resistance (R=
V
/I).
5. Repeat the experiment with nichrome wire, and then record the voltmeter
and ammeter readings to calculate the resistance.
A
V
199
Ohm’s law
A German physicist, George Simon Ohm was a physics teacher. In 1826 he published a book
containing details of the experiments he made to investigate the relationship between the current
passing through the wire and the potential difference across the supply at a constant temperature.
The experiments that he carried out had the results equivalent to the ones shown in Table 11.4.
Table 11.4
Number of
cells
Potential difference
(voltage) V
Current I (A)
Voltage
Current
V
I
1
1.5
0.1
1.5
0.1
15
2
3
0.2
3
0.2
15
3
4.5
0.3
4.5
0.3
15
4
6
0.4
6
0.4
15
From Table 11.4, it has been noted that current increases as potential difference increases.
This relation is called Ohm’s Law.
Ohm’s law states that the current in a conductor is directly proportional to the potential
difference between its ends, at constant temperature.
I α V
V = I x constant
Exercise 11.3
In your groups,
1. Explain why the conductivity of different metals is different.
2. Discuss with the help of a circuit diagram, how you would measure
the resistance of a resistor.
3. Describe an experiment that you would carry out to find the length of
a nichrome wire which can give a resistance of 0.8 Ω.
200
V = constant
I
Constant is called Resistance, R
Therefore, V = R
I
or V = IR
or I = V
R
Whereby:
I is current in Amperes (A)
V is potential difference in volts (V)
R is resistance in Ohms (
)
V
I
R
201
Verifying Ohm’s law
DISCUSSIONS
1. Calculate the resistance for each number of cells.
2. What have you noticed about the results in question 1? Give a reason for your answer.
3. Do the results in Experiment 11.5 verify the Ohm’s law? Give a reason for your answer.
Experiment 11.5
AIM: To verify Ohm’s law
MATERIALS: Cells, switch, ammeter, voltmeter, resistor (a nichrome wire) and
connecting wires.
PROCEDURE:
1. Set up the apparatus as shown in Figure 11.15.
2. Close the switch and take the voltmeter and ammeter readings.
3. Repeat the experiment with 2, 3 and 4 cells. Take the voltmeter and ammeter
readings for each number of cells in Table 11.5.
Number of
cells
Voltmeter
reading (V)
Ammeter
reading (A)
Resistance
(Ω)
Voltage
Current
1
2
3
4
Table 11.5
Figure 11.15
A
R
V
202
4. Plot a graph of voltmeter reading (V) against the current reading (A).
5. Describe the shape of the graph in question 4.
6. Does the graph in question 4 verify the Ohm’s law? Give a reason for your answer.
Ohm’s law is applied only to some materials (metals and some alloys). Materials that obey
Ohm’s law, and hence have a constant resistance over a wide range of voltages, are said to be
ohmic. Ohmic materials have a linear current–voltage relationship over a large range of applied
voltages as shown in Figure 11.16.
Materials having resistance that changes with voltage or current are nonohmic. Nonohmic
materials have a nonlinear current–voltage relationship as shown in Figure 11.17.
One common semiconducting device that is nonohmic is the diode.
Figure 11.16 current-voltage-curve for an ohmic material
Figure 11.17 current-voltage curve for a nonohmic material
203
Measuring resistance using an ohmmeter
A resistor is a device that causes resistance.
It is used to reduce the amount of current flowing in the circuit. The symbol for a resistor is
shown below:
In Experiment 11.5, the resistance of the resistor is measured using the Ohm’s law.
Using Ohm’s law, you connect a voltmeter across a resistor and an ammeter in series in the
circuit. The resistance of a resistor is found by dividing the voltmeter reading by the ammeter
reading.
Resistance = Voltmeter reading
Ammeter reading
R = V
I
The resistance of a resistor can also be found by using an ohmmeter. An ohmmeter is an
instrument which is used to measure resistance of a resistor. Using an ohmmeter, you connect
one terminal of an ohmmeter to one side of a resistor and the other terminal of an ohmmeter to
the other side of a resistor. Take the reading of resistance on an ohmmeter.
Calculating resistance using Ohm’s law
Worked examples
1. A potential difference of 30 V is needed to make a current of 5 A to flow through a wire.
Calculate the resistance of the wire.
Solution
V = 30 V I = 5 A R =?
R = V
I
= 30 V
5 A
R = 6 Ω
204
2. Calculate the voltage across a 10
resistor carrying a current of 0.5 A.
Solution
R = 10
I = 0.5 A V =?
V = IR (Ohm’s law)
V = 0.5A x 10
V = 5 V
3. Calculate the current flowing through a 2
resistor when a potential difference of 3 V is
applied across it
Solution
V = 3 V R = 2
I =?
V = IR
I = V
R
I = 3 V
2
I = 1.5A
Exercise 11.4
In your groups, answer the following questions:
1. Explain how you can verify Ohm’s law experimentally.
2. A resistor has a resistance of about 200 Ω. Calculate the potential difference
required to produce a current of 2 A.
3. A current of 300 mA flows through a 3 kΩ resistor. Calculate the potential
difference across the resistor.
4. Explain why temperature must always be constant when verifying Ohm’s law.
5. You are employed as a technician at Flames Television and you have been asked to
find the resistance of an electronic component.
a. Draw a circuit diagram that you can use, showing all materials that you need.
b. Show how you can use the results obtained from your set up to find the resistance
of a component.
205
Finding resistance of resistors using colour codes and standard
notation
Colour codes
The resistance in ohms can be marked on the resistor using colours. This method is called colour
coding or resistance coding.
Each colour has its own standard notation (digit).
Resistors are colour coded with four or five bands to indicate their resistance.
Figure 11.19 shows a resistor that is colour coded.
Explanation of bands on a colour coded resistor
Figure 11.19 4 Band Resistor colour code layout
Figure 11.18 resistor colour codes
206
The last band always gives tolerance. Tolerance is the extent to which the actual value of the
resistance can vary.
The following are the values of tolerance:
Colour Tolerance
Gold ± 5%
Silver ± 10%
No colour ± 20%
The band which is second from the last band gives the number of noughts ( zeros) called
multiplier.
Other bands give the digits.
The following are colour codes:
Colour digit
Black 0
Brown 1
Red 2
Orange 3
Yellow 4
Green 5
Blue 6
Violet 7
Grey 8
White 9
Worked examples
Find the resistance of each of the following resistors.
1.
Solution
Last band is silver, therefore tolerance = ±10%.
Second from last band is Brown: therefore, number of noughts =1(1 zero).
The first band is Red. Therefore, the first digit is 2.
The second band is Violet. Therefore, the second digit is 7.
The value of the resistance will be written as:
Red Violet Brown Silver
Figure 11.20
207
Red Violet Brown Tolerance
2 7 0 ±10%
Resistance = 270 ±10%Ω
2.
Solution
Last band is Gold, tolerance = ±5%
Second from last band is black, noughts = none
First band is Brown, first digit = 1.
Second band is Black, second digit = 0
Third band is White, third digit = 9
The value of the resistance will be written as:
Brown Black White Black Gold
1 0 9 - ± 5%
Resistance = 109 ± 5% Ω
Standard notation
The resistors have numbers and letters printed on them.
Examples
Brown Black White Black Gold
Figure 11.21
18Ω 18KΩ
0.18Ω 0.18 KΩ
1.8Ω 1.8KΩ
Figure 11.22 resistance code standard notation
18 R
R18
1R8
18K
K18
1K8
208
The letters are used as tolerance.
Examples
Internal resistance of a cell
When a voltmeter is connected across a 1.5 V cell, it reads 1.5 V. This shows that chemical
action within the cell causes an e.m.f. of 1.5 V.
When the cell is connected in a circuit and supplies current, the voltmeter reads 1.2 V. This is the
potential difference (p.d) in the circuit. The reading on the voltmeter has dropped because the
cell has resistance, like other components. This resistance is called internal resistance, r.
Internal resistance (r) is the resistance of a cell or battery to the current it causes. It is the
resistance of the connections in the cell and some chemical effects e.g. polarization. The internal
resistance is usually low, about 0.5Ω or so.
Internal resistance = “lost in voltage”
current
r = v
I
Letter Tolerance
F ± 1%
G ± 2%
J ± 5%
K ± 10%
M ± 20%
18±2% Ω
0.18±5% Ω
1.8±20% Ω
Figure 11.23 resistance code standard notation
18RG
R18J
1R8M
209
Lost in voltage, v = Ir
Electromotive force, E = potential difference, V + lost in voltage, Ir
E = V + Ir
Worked example
The e.m.f. across the terminals of a cell is 3.0 V. If the p.d. across the cell is 2.5 V and the
current flowing is 2 A, calculate the internal resistance of a cell.
Solution
E = 3 V V = 2.5 V I = 2 A r =?
Either: E = V + Ir
3 V = 2.5 V + (2 A x r)
3 V – 2.5 V = 2 A x r
0.5 V = 2 A x r
r = 0.5 V
2 A
r = 0.25 Ω
Or: Internal resistance = “lost in voltage”
current
lost in voltage, v = E – V
= 3.0 V – 2.5 V
v = 0.5 V
r = V
I
r = 0.5 V
2 A
r = 0.25 Ω
2. A battery has an e.m.f. of 12 V and an internal resistance of 0.6 Ω. What is the p.d. across
its terminals when it is supplying a current of 5 A?
Solution
E = 12 V V=? r = 0.6 Ω I = 5 A
E = V + Ir
V = E – Ir
V = 12V – (5 A x 0.6 Ω)
210
V = 12 V – 3 V
V = 9 V
11.4 Electric circuits
An electric circuit is a conducting path in which electrons flow or electric current takes place.
An electric circuit can consist of a cell or battery, connecting wires, bulb, resistors, ammeter and
voltmeter.
Exercise 11.5
In your groups answer the following questions:
1. Define the internal resistance.
2. Calculate the resistance of the following resistors:
a.
b.
3.
The e.m.f. across the terminals of a cell is 5.0 V. If the p.d. across the cell is
3.7 V and the current flowing is 2.5 A, calculate the internal resistance of a cell.
4. A battery has an e.m.f. of 1.5 V and an internal resistance of 0.2 Ω. What is the
p.d. across its terminals when it is supplying a current of 2 A?
211
Circuits are grouped into two:
a. Series circuit
A series circuit is a circuit in which all the components are connected in one line. A
series circuit has one conducting path.
b. Parallel circuit
A parallel circuit is a circuit in which components are connected in branches. A parallel
circuit has more than one conducting path.
Figure 11.26 series circuit
R1
A
R2
Figure 11.27 parallel circuit
A
R1
R2
212
The effect of resistors in series and parallel circuits.
Experiment 11.6
AIM: To investigate the effect of resistors in series and parallel circuits
MATERIALS: Cells, switch, ammeter, voltmeter, 3 nichrome wires (for any 3 resistors)
and connecting wires.
PROCEDURE:
1. Set up the apparatus as shown in Figure 11.28.
2. Close the switch, and then record the ammeter and voltmeter readings in the table.
Voltmeter reading
(V)
Ammeter reading
(A)
Resistance(Ω)
Table 11.6
3. Calculate the total resistance in the circuit by using a formula:
R = V
I
4. Repeat the experiment by connecting the resistors in a parallel circuit as shown in
Figure 11.29.
213
RESULT
When the resistors are connected in a series circuit the resistance in the circuit is higher than the
resistance when they are connected in a parallel circuit.
EXPLANATION/CONCLUSION
Experiment 11.6, shows that the total resistance of the resistors connected in a series circuit is
greater than the total resistance of the resistors connected in a parallel circuit.
Current in series and parallel circuits
Current in series circuit
Current across each and every component in series circuit is the same. This current is the same as
the current from the supply.
In this circuit, A1 = A2 = A3
Current in parallel circuit
When components are connected in a parallel circuit, the sum of the currents in a parallel circuit
equals the current in series (main) circuit.
Figure 11.30 current flowing in series circuit
A1
R1
R2
A2
A3
Figure 11.31 current flowing in a parallel circuit.
A1
R1
R2
A2
A3
214
Sum of the current in parallel circuit = current in the main circuit
A2 + A3 = A1
Voltage in series and parallel circuit
Voltage in series circuit
The sum of the voltages across the components connected in series circuit equal the voltage or
Pd from the supply.
Figure 11.32 voltages in series circuit.
From the circuit shown in Figure 11.32:
V2 + V3 = V1
R1
R2
V1
V3
V2
215
Voltage in a parallel circuit
Voltage across each component connected in parallel circuit is the same and equal to the supply
voltage.
Figure 11.33 voltages in parallel circuit.
From the above circuit:
V1 = V2 = V3
Net resistance of resistors connected in series and parallel
circuits
Net resistance of resistors connected in series
If resistors are connected in series, they give a higher resistance than any one of the resistors by
itself because the effect is the same as joining resistance wires together to form a longer wire.
The resistance of the resistance wire increases with length.
To find the total resistance of the resistors connected in series:
R1
R2
V1
V2
V3
216
Figure 11.34 resistors connected in series.
Current through R1, R2 and R3 is the same current I since the resistors are in a series circuit.
The sum of the voltage across R1, R2, and R3 equals the voltage from the source (VT)
VT = V1 + V2 +V3………………… (i)
But V = IR
V1 = IR1
V2 = IR2
V3 = IR3
VT = IRT
Substituting in for the values of V in equation (i):
IRT = IR1 + IR2 + IR3
Dividing throughout by I, the final equation becomes:
RT = R1 + R2 + R3
Therefore, total resistance of resistors in series circuit is found by the formula:
RT = R1 + R2 + R3 + …………….. Rn
For example: The resistors 3Ω, 2Ω and 4Ω will have a total (net) resistance of 9Ω as shown
below:
RT = 3Ω + 2Ω + 4Ω
RT = 9Ω
Net resistance of resistors connected in parallel
217
If resistors are combined in parallel they give a lower resistance than any one of the resistors by
itself because the effect is the same as connecting a thick resistance wire. The resistance of a
resistance wire decreases with an increase in thickness of the wire. Therefore, combined
resistance is less than the resistance of the smallest individual resistor.
To find the combined resistance of the resistors connected in parallel:
Figure 11.35 resistors connected in parallel
Voltage in parallel circuit is the same V.
The sum of the current in parallel circuit equals the current from the supply.
IT = I
1
+ I
2
+ I
3
……………………… (i)
But I = V
R
I
1
= V
R1
I
2
= V
R2
I
3
= V
R3
IT = V
RT
Substituting for the values of I in equation (i):
V = V + V + V
RT R1 R2 R3
Dividing throughout by V, the equation becomes:
1 = 1 + 1 + 1
RT R1 R2 R3
3Ω
2Ω
4Ω
R1
R3
R2
218
Therefore, the total resistance of resistors in parallel circuit is given by the formula:
1 = 1 + 1 + 1 + ….... 1
RT R1 R2 R3 Rn
For example: If the 3Ω, 2Ω and 4Ω resistors are connected in parallel circuit, the total (net)
resistance can be worked out as follows:
1 = 1 + 1 + 1
RT R1 R2 R3
1 = 1 + 1 + 1
RT 3Ω 2Ω 4Ω
1 = 4 + 6 + 3
RT 12
1 = 13
RT 12
RT = 12
13
RT = 0.92Ω
When two resistors are connected in parallel their effective resistance can be worked out using a
formula:
RT = the product of their resistances
the sum of their resistances
RT = R1 x R2
R1 + R2
For example: 2Ω and 4Ω resistors are connected in a parallel circuit as shown in
219
Figure 11.36 below:
RT = R1 x R2
R1 + R2
RT = 2Ω x 4Ω
2Ω + 4Ω
RT = 8
6
RT = 1.33Ω
Circuit problems
Worked examples
1. Calculate the total resistance in each of the following:
Figure 11.36
2Ω
4Ω
(a) (b)
(c)
(d)
Figure 11.37
4Ω
2Ω
2Ω
3Ω
4Ω
6Ω
2Ω
8Ω
2Ω
4Ω
6Ω
220
Solution
a. Resistors in series,
RT = R1 + R2
RT = 4
+ 2
RT = 6
b. Resistors in parallel circuit;
1 = 1 + 1 + 1 + 1
RT R1 R2 R3 R4
1 = 1 + 1 + 1 + 1
RT 2
3
4
6
1 = 6 + 4 + 3 + 2
RT 12
1 = 15
RT 12
1 = 12 Ω
RT 15
RT = 0.8 Ω
c. Two resistors in parallel,
RT = R1 x R2
R1 + R2
RT = 2Ω x 8Ω
2Ω + 8Ω
Therefore, RT = 1.6 Ω
d. Two resistors in parallel and 1 resistor in series,
RT in parallel and R in series
R1 R2 + R
R1 + R2
RT in parallel,
RT = R1 R2
R1 + R2
221
RT = 2Ω x 4Ω
2Ω + 4Ω
RT in parallel = 1.3 Ω
Total resistance, RT in the circuit = RT in parallel + R = 1.3
Ω
+ 6
Ω
RT in the circuit = 7.3 Ω
2. The resistor with unknown resistance is connected in parallel to an 8
resistor.
Calculate the value of the unknown resistor if the effective resistance is 4
.
Solution
R1 =? R2 = 8
RT = 4
RT = R1 x R2
R1 + R2
4 Ω = 8Ω x R1
R1 + 8Ω
4 (R1 + 8) = 8 x R1
4R1 + 32 = 8R1
32 = 8R1 – 4R1
32 = 4R1
R1 = 32
4
R1 = 8 Ω
3. Figure 11.38 is a simple series circuit.
Calculate:
a. The total current in the circuit.
b. The voltage across a 2Ω resistor.
222
Solutions
a. IT = V
RT
V = 12V
RT = R1 + R2
= 2
+ 4
RT = 6
IT = 12V
6Ω
IT = 2A
b. V across a 2
Ω resistor
V = IR
I = 2A (current is the same in series circuit)
R = 4Ω
V = 2A x 4Ω
Therefore, V = 8V
4. A 2Ω and 4Ω resistors are connected in parallel and a 6Ω resistor is connected in series with
them. A voltage across the battery is 12V.
Find:
a. The total current in the circuit.
b. The current in the 4Ω resistor.
12 V
Figure 11.38
2Ω
4Ω
223
Solutions
a. IT = V
RT
V = 12V
RT = R1 x R2 + R3
R1 + R2
RT = 2 x 4 + 6
2 + 4
RT = 1.3 + 6
RT = 7.6Ω
IT = 12V
7.6Ω
IT = 1.6A
b. I in the 4Ω resistor
First, let us find voltage across a 4Ω resistor
V across (4Ω and 2Ω) + V across 6Ω = 12V
V across 6Ω = ITR
V across 6Ω = 1.6A x 6Ω
V across 6Ω resistor = 9.6 V
V across a 4Ω resistor = 12 V – 9.6 V = 2.4 V
I in a 4Ω resistor = V
R
12 V
Figure 11.39
6Ω
2Ω
4Ω
224
= 2.4 V
4Ω
I in a 4Ω resistor = 0.6 A
Exercise 11.6
In your groups, answer the following questions:
1. Calculate the combined resistance in each case:
2. In Figure 11.41, calculate:
a. the current in the main circuit
b. the p.d. across a 5 Ω resistor
c. the current through the 10 Ω resistor
a. b.
c.
Figure 11.40
3Ω
3 Ω
3 Ω
3 Ω
2 Ω
3 Ω
4 Ω
4 Ω
30 V
Figure 11.41
10 Ω
4 Ω
5 Ω
225
11.5 Electric power and energy
Electric power
Power is the rate of doing work, or it is the electrical energy transferred per unit time, or it is the
rate at which energy is produced.
Power = work done
Time taken
OR Power = energy used
Time taken
In an electric circuit, power is provided by the cell or battery. Amount of power generated by a
cell or battery is the product of voltage and current flowing in the circuit.
Electrical power = voltage x current
P = IV ………………(i)
Power dissipated in a resistor of resistance R
Since P = IV
But I = V (Ohm’s law)
R
Substitute V for I in equation ….(i)
R
P = V x V
R
Therefore, P = V
2
……(ii)
R
V = IR (Ohm’s law)
Substitute IR for V in equation (i)
P = I x R x I
Therefore, P = I
2
R…….(iii)
226
The three equations used for calculating electrical power are:
P = VI……………(i)
P = V
2
……………(ii)
R
P = I
2
R…………..(iii)
Power is measured in watts. SI unit of power is the watt (w).
Other units of power are:
1 kilowatt (Kw) = 1000W = 10
3
W
1 megawatt (Mw) = 1 000 000W = 10
6
W
Electrical power calculations
Worked examples
1. In an electric circuit, the pd across the battery is 3 V and the current supplied is 2 A.
Calculate the power supplied by a battery in the circuit.
Solution
V = 3V I = 2A P =?
P = IV
P = 2A x 3V
P = 6 W
2. Find the resistance of a 60 W electric lamp if it uses 240V.
Solution
V = 240 V P = 60 W R =?
P = V
2
R
R = V
2
P
R = 240
2
60
R = 960 Ω
227
3. Calculate the total power dissipated in the resistors shown in the diagram.
Solution
EITHER
V = 12V
RT = 10 Ω + 20 Ω = 30 Ω
P = V
2
RT
P = 12
2
30
P = 4.8 W
OR
V = 12V RT = 30 Ω
IT = V
RT
I = 12V
30 Ω
I = 0.4A
P = I
2
RT
P = 0.4
2
x 30 Ω
P = 4.8 W
12 V
Figure 11.42
10 Ω
20 Ω
228
Electric energy
Since power = Energy used
Time taken
P = E
t
Therefore, energy = power x time
E = P x t
But P = IV, P = I
2
R, P = V
2
R
Therefore, electrical energy will have the following equations:
Exercise 11.7
In your groups, answer the following questions:
1.
The power rating of a heating coil is 50 W. If a coil is connected to a supply
of 240 V, calculate the resistance of the coil and the current through it.
2. A current of 2 A flows through a 4 Ω resistor. What is the power dissipated
across the resistor?
3. In the circuit diagram shown in Figure 11.43, calculate the power supplied
by a 1.5 V cell.
4. Elimat hair salon uses a hair dryer rated 240 V, 200 W. Explain the meaning
of 240 V, 200 W.
1.5 V
0.25 A
Figure 11.43
A
229
E = P x t……………(i)
E = IVt…………….(ii)
E = I
2
Rt……………(iii)
E = V
2
t .....................(iv)
R
Energy is measured in Joules (J).
1 Kilojoule (KJ) = 1000J = 10
3
J
Worked Examples
1. An electric bulb is rated at 100 W. Calculate the energy used by the bulb in
a. 5 seconds
b. 3 hours
Solution
P = 100 W V = 220V
a. Energy used in 5 s
t = 5
E = Pt
E = 100 w x 5 s
E = 500 J
b. Energy used in 3 hrs
t = 3 hrs = 10800 s
E = p x t
E = 100 W x 10800 s
E = 1080000 J or 1.08
6
J or 1080 KJ
2. Calculate the energy used by a resistor after 10 s
I = 0.2A R = 6 Ω t = 10 s
E = I
2
Rt
0.2 A
Figure 11.44
A
6 Ω
230
E = 0.2
2
A x 6 Ω x 10 s
E= 2.4 J
Cost of electricity
Electricity is supplied by the local Electricity Board. For example, in Malawi electricity is
supplied by the Electricity Supply Corporation of Malawi Ltd (ESCOM). The Electricity board
charges electricity in form of electrical energy used by appliances. The electrical energy is
measured in kilowatt – hours by the electric energy meter.
1 kilowatt – hour is sold as 1 unit of electrical energy.
A kilowatt hour or unit of electricity is the electrical energy supplied in 1 hour to an
appliance whose power is 1kw.
Total cost of electricity = kilowatt – hours x cost per kilowatt – hour
Worked examples
1. An electrical appliance is rated at 200 W. If electrical energy costs K28.00 per kWh,
what is the cost of using this heater for 8 hours at its maximum power?
Solution
P = 200 W = 0.2 kW t = 8hrs 1 kWh = K28.00
Electrical energy = p x t = 0.2 kW x 8 h
= 1.6 kWh
Figure 11.45 Electricity board’s meter
231
Total cost of electricity = 1.6 kWh x K28.00
Total cost of electricity = K44.80
2. Figure 11.46 shows appliances used in a house
If electrical energy costs K27.90 per unit (1kWh), calculate the total cost of using a fan and two
bulbs for three hours per day for one week.
Solution
Total power = 1.2 kW + 0.6 kW + 0.6 kW = 2.4 kW
t = 3 h x 7 = 21 hours
E = p x t
Energy = 2.4 kW x 21 h
Energy = 50.4 kWh
Total cost = electrical energy x cost of 1 kWh
= 50.4 kWh x K27.90
Total cost = K1406.16
A.c. supply
Figure 11.46
0.6 kW
0.6 kW
Fun 1.2 kw
232
Interpreting the electricity bill
Table 11.7 shows an electricity bill prepared by Magesi Electricity Supply Company of Malawi
(MESCOM).
Table 11.7 An electricity bill
From the bill in table 11.7:
Units used by Mr Grezo = The difference in meter readings in kWh
= 46839 – 46523
Units used = 316.00 kWh
If 1 kWh = K29.00
Cost for 316.00 kWh = 316.00 x K29.00 = K8 848.00
SURTAX= 16.50% of K8 848.00
Magesi Electricity Supply Company of Malawi (MESCOM), P O Box 1340, Limbe.
Electricity account covering period of approximately one month preceding date of meter reading
DATE
DETAILS OF TRANSACTIONS
AMOUNT
(MKW)
BALANCE (MKW)
01/06
10/06
30/06
SURTAX
Balance brought forward
Receipts
Previous Current consumption
46523 46839 316.00
16.50% on 8 848.00
Total current bill
8 848.00
1 459.92
10 307.92
10 317.54
10 000.00-
10 307.92
The monthly total is for JUNE 2014 and is payable 15/07/2014
AMOUNT DUE
10 625.46
Mr Grezo
P.O Box 55
Blantyre
A/C Number: 200903775
Tariff : M11
Reference: : 104
233
= 16.50 x K8 848.00
100
SURTAX = K1 459.92
Mr Grezo’s total monthly bill = K8848.00 + K1459.92 = K10307.92
Mr Grezo’s amount due = (Total monthly bill + previous month’s balance) – amount paid
Mr Grezo’s amount due = (K10 307.92 + K10 317.54) – K10 000
Mr Grezo’s amount due for the month of June = K10 625.46
Power of the heating element
The current has the heating effect when it flows through a resistive material e.g. a coil of an
electric heater or cooker. The power of the heating element can be found in the following
formulas:
P = VI……………(i)
P = V
2
……………(ii)
R
P = I
2
R…………..(iii)
Worked example
An electric cooker is connected to a 240 V mains. If the resistance of its coils is 15Ω, calculate
its power rating.
Solution
V = 240 V R = 15Ω
P = V
2
R
= 240
2
15
P = 3840 W or 3.840 kW
Heating elements work under different currents and voltages. Hence, they have different power
ratings.
Energy transfer
The electrical energy in heating elements is released as heat energy. Therefore, heat energy lost
from a heating element is gained by the surrounding material, e.g. electric heater. In this case, we
say electric current has a heating effect. This is noticed by an increase in temperature.
234
If there is no energy loss then,
Electrical energy = Heat energy
Electrical energy = power x time
E = p x t
Heat energy = mass x specific heat capacity x change in temperature
HE = m x c x ∆T
The equations can be related as follows:
P x t = m x c x ∆T………………...(i)
IVt = m x c x ∆T ………………(ii)
I
2
Rt = m x c x ∆T ……………….(iii)
V
2
t = m x c x ∆T ……………….(iv)
R
Worked example
The immersion heater is used to heat water in a bath. If a heater rated at 3.6W is connected to
240V main supply, calculate
a. The resistance of the heating element.
b. The time taken for 2 kg of water in a bath to raise its temperature from 20
0
C to 25
0
C.
(SHC of water is 4200J/kg
0
C).
Solution
a. Resistance of the heating element
P = 3.6 kW = 3600 W
V = 240 V
P = V
2
R
R = V
2
P
R = 240
2
3600
R = 16 Ω
b. Time taken for 2 kg of water to raise its temperature from 20
0
C to 25
0
C
P = 3600 W ∆t = (25
o
C – 20
o
C) = 5
0
C m = 2 kg
c = 4 200/kg
o
C
Pt = m x c x ∆T
235
3600 x t = 2 x 5 x 4200
t = 2 x 5 x 4200
3600
t = 11.7 s
The electrical hazards
Although electricity is very useful, it can be dangerous when it is not used safely. Therefore,
electricity can be hazardous.
The major hazards associated with electricity are:
1. Electric shock
An electric shock is the passing of electric current through the body. The body becomes part of
the electric circuit.
An electric shock can happen in the following situations:
a. When the body comes into contact with both wires (live and neutral wires) of an electric
circuit.
b. When the body comes into contact with a metallic part that has become live (energized)
through contact with an electrical conductor.
Exercise 11.8
In your groups, answer the following questions:
1. A 1.5 kW refrigerator is switched on for 5 hours. Calculate the electrical energy in the fan
in joules.
2. A heating coil is connected to a 240 V supply and a current of 10 A flows in it.
If the coil is used to heat 0.1 kg of water in 100 s, calculate the temperature increase of
water.
3. If electrical energy costs K28.00 per unit, calculate the cost of:
a. leaving a 5kW heater witched on for 6 hours.
b. leaving a 100 W bulb switched on for 1 day.
4. An immersion heater rated 2200 W is used to change the temperature of water from 25
0
C
to 80
0
C in 1 hour. Calculate the mass of water heated. (Specific heat capacity of water is
4200 J/kg
0
C.
236
Electric shock depends on a number of factors such as the pathway through the body, the amount
of current, the length of time of the exposure, wetness of the skin and presence of water (if the
area is damp or dry).
The effect of electric shock may range from a slight tingle to severe burns or to cardiac arrest.
Table 11.8 shows the general relationship between the degree of injury and amount of current.
Table 11.8 relationship between the degree of injury and the amount of current
Amount of current (mA)
Degree of injury
1
Perception level
5
Slight shock felt, not painful but disturbing
6-30
Painful shock
50-150
Extreme pain, respiratory arrest, sense muscular
contraction
1000- 4300
Ventricular fibrillation
10000+
Cardiac arrest, severe burns and probable death
2. Overheating and fire
When high current flows through the cable or appliance, there will be overheating and fire.
High currents can be caused in the following ways:
a. Short circuit
A short circuit is the accidental touching of a live wire and a neutral wire.
At the point of short circuit, the resistance becomes very low. This allows high current to flow
through the circuit. The high current can cause overheating and fire.
b. Overloading
Overloading happens when a lot of appliances are connected on one surface and all the
appliances are switched on at the same time e.g. on an extension. The appliances take more
current to the surface. This high current can cause overheating and fire.
Preventing electrical hazards
There are various ways of protecting people from the hazards caused by electricity.
237
Basic precautions
Follow some basic precautions as listed below:
Inspect wiring of equipment before each use. Replace damaged or frayed electrical cords
immediately.
Use safe work practices every time electrical equipment is used.
Know the location and how to operate switches or circuit breakers. Use these devices
(switch or circuit breaker) to shut off equipment in the event of a fire or electrocution.
Limit the use of extension cords; use them only for temporary operations.
Use plugs that are equipped with circuit breakers or fuses.
Place exposed electrical conductors behind the shields.
Minimise the potential for water or chemical spills on or near electrical equipment. All
electrical cords should have sufficient insulation to prevent direct contact with wires. It is
very important to check all cords before each use, since corrosive chemicals or solvent
vapours may corrode the insulation.
Circuit protection devices
Circuit protection devices are designed to automatically limit or shut off the flow of electricity in
the event of a ground-fault, overloading or short circuit in the wiring system. The circuit
protection devices are fuses and circuit breakers.
Fuses and circuit breakers prevent overheating of wires and components.
Fuse: A fuse is used to control the amount of current flowing in the circuit. If there is high
current flowing in the circuit by accident, the fuse melts. This breaks the circuit and stops the
flow of current. Hence the wire and the appliances are protected from high current, overheating
and fire.
Symbol for a fuse
A fuse is found in a three-pin plug.
238
Three-pin plug
A three-pin plug is used to connect appliances to the mains. Three-pin plugs are used because
they are safe since they have a fuse.
Live wire
A live wire carries ac current and voltage to the appliance
It has alternating voltage that moves to +240V then to -240V, making the
alternating current which flows backwards and forwards through the circuit.
It gives electrical shock when touched.
Neutral wire
A neutral wire acts as a returning path of ac current and voltage.
It has a potential difference of O V
It does not give an electrical shock when touched.
Earth wire
It is a safety wire; it prevents users from getting electrical shock.
An earth wire is grounded at one end and connected to a metallic part of an
appliance at the other end. When faulty current flows, a metallic appliance
becomes live. The earth wire conducts the faulty current to the ground. Hence
preventing users from electric shock.
Fuse
A fuse is a safety device that prevents cables and appliances from carrying high current.
High current flowing in cables or appliances can cause overheating and fire.
When the current flowing in a cable or appliance is more than required amount, the fuse
melts and breaks the circuit.
It is connected to the live wire, since it is the live wire that carries current.
Figure 11.47 three-pin plug
239
Fuse rating
Fuse rating is the maximum amount of current that a fuse can allow to pass through before it
melts.
To find fuse rating of a fuse:
Fuse rating = power supplied
Voltage from the supply
Worked example
An appliance rated at 60 W, uses a voltage of 35 V. Calculate the fuse rating of an appliance.
Fuse rating = Power = 60 W
Voltage 35 V
Fuse rating = 1.7 A
But the value of the fuse rating should not be exactly 1.7 A. There must be an allowance to allow
maximum current to flow.
Circuit breaker: This is an automatic switch which if the current rises over a specified value,
the electromagnet pulls the contacts apart, thereby breaking the circuit. The reset button is to rest
everything. It works like a fuse, but it is better because it can be reset.
Exercise 11.8
In your groups, answer the following questions:
1. Explain how each of the following can cause electrical hazards:
a. Damp condition
b. High current
c. Damaged insulation
2. Explain how a fuse works.
3. In a three-pin plug, which wire
a. has a blue covering
b. has a yellow and green covering
c. is a safety wire?
4. What is the function of a circuit breaker?
240
Summary
An electric current is the flow of electric charges (electrons) from the negative side of an electric
field to the positive side.
The SI unit of electric current is Ampere (A).
An electric current is also defined as the rate at which electric charge flows.
I = Q
t
Potential difference is the difference in potential between two points, equal to the energy change
when a unit electric charge moves from one place to another in an electric field. The SI unit of
potential difference is the Volt (V).
Electromotive force (EMF) is the maximum potential difference across a cell or battery when it
is not in a circuit and not supplying current (when I= 0 A).
Electrical resistance is the opposition to the flow of electrons in a wire or a circuit. The SI unit of
electrical resistance is Ohm (Ω).
Factors that affect electrical resistance of a wire are length of a wire, temperature, cross sectional
area and nature of the material.
The total resistance of the resistors in series is found as:
RT = R1 + R2 + R3 + ……Rn
The total resistance of the resistors in parallel is found as:
Electric power is the rate at which electrical energy is transferred.
P = E
t
P = VI, P = I
2
R, P = V
2
R
Electrical energy = power x time
E = P x t, E = VIt, E = I
2
Rt E = V
2
t
R
Electrical hazards are electric shocks, overheating and fire.
1 = 1 1 1 1
RT R1 R2 R3 R4
241
Student assessment
1. Define
a. Electric current.
b. Electric resistance.
2. A charge of 100 C flows in the circuit for 2 hours. Calculate the amount of current
flowing in the circuit.
3. State Ohm’s law.
4. Describe the experiment that you would carry out to verify Ohm’s law.
5. To verify Ohm’s law for a piece of metal wire, a student obtained and came up with the
following data.
Table 11.8
Voltage(V)
1.5
1.8
2.1
2.4
2.7
3
Current(A)
0.75
0.9
1.05
1.2
1.35
1.5
a. Plot a graph of voltage against current.
b. Using your graph find the voltage when current is 1A.
c. If the wire obeys ohm’s law, what do you notice about the values (V/A)
in a graph?
6. State the difference between potential difference (PD) and electromotive force (Emf).
7. In each of the following circuits, find the total or effective resistance.
8. In Figure 11.49, calculate:
a. The total current in the circuit.
a. b.
Figure 11.48
3Ω
2Ω
6Ω
5Ω
4Ω
2Ω
242
b.
Voltage across a 3Ω resistor.
9. Figure 11.50 below is a diagram of an electric circuit.
a. Calculate the total resistance under the following conditions:
i. S open
ii.
S closed
b. Calculate the total current in the circuit under the following conditions:
i. S open
ii.
S closed
10. Define colour coding.
11. Explain why colour coding is used in resistors found in electronic appliances.
12. Figure 11.51 below shows a resistor. Use it to answer the questions that follow.
Figure 11.49
Figure 11.50
Red Orange Brown Gold
1.5 V
3Ω
3Ω
3Ω
S
243
a. Work out the resistance of a resistor.
b. If a resistor is connected in the circuit which operates from a supply of 240V,
what current will flow in the circuit?
13. Describe the experiment that you would carry out to show that electrical resistance of a
wire varies directly proportional to its length.
14. Define electrical power.
15. The power rating of a heating coil is 50 W. If a coil is connected to a supply of 240 V.
Calculate the resistance of the coil.
16. A current of 2A flows through a 4 Ω resistor. What is the power dissipated across the
resistor?
17. In the circuit diagram shown in Figure 11.52, calculate the power supplied by a 3 V
battery.
18. Explain why it is encouraged to transmit power at very high voltage and very low current.
19. A 2kW power is fed to a transmission cable of resistance 2 ohms. How much power is
wasted in the cable if power is transmitted at a current of
a. 3A?
b. 1A?
Figure 11.52
3V
10 Ω
244
20.
An electric fuse is designed to operate at 10 A from a 240 V supply. Calculate the
electrical energy supplied to a heater in 2 hours.
21.
A heating coil is immersed in 0.5 Kg of water. The coil is connected to a 15 V supply
and a current of 2 A flows for 140 seconds. Calculate the temperature increase of water
(SHC of water = 4200J/Kg
0
C).
22.
The following appliances are connected in a house.
Item Power
Fan 2.2Kw
Refrigerator 200W
Bulb 100W
Cooker 3KW
a. Calculate the total power taken from the supply if all items are running.
b. The appliances are connected to a 240V supply. Find
i. current in a bulb
ii. Resistance in coils of a cooker.
c. If the cost of electricity is K27.00 per kWh,
i. Define a kilowatt-hour.
ii. Calculate the total cost of running all the items for 5 hours.
23. Explain how you can relate electrical energy and heat energy.
24. Figure 11.53 below shows the inside of an electric plug. Use it to answer the questions
that follow.
a.
Figure 11.53
B
A
C
D
245
a. Name the parts labeled A, B, C and D.
b. State the functions of parts A, B, C and D.
c. If the three-pin plug is used for a 240V, 2 kW appliance, work out the fuse rating of the
fuse in the plug.
CHAPTER 12
246
Oscillations and waves
Objectives
At the end of chapter 12, you must be able to:
Explain oscillation in relation to a
pendulum or hanging mass on a spring
Describe a wave
Differentiate between a transverse wave
and longitudinal wave
Describe wave properties
Apply the wave equation in solving
problems
12.1 Oscillations or Vibrations
Oscillations or vibrations are complete upward or downward movements of an object about its
fixed position (rest position or equilibrium position).
Oscillations or vibrations can also be defined as complete to and fro movements of an object
about its fixed position (rest position or equilibrium position).
Oscillations are produced by the vibrating systems.
247
Examples of vibrating systems are vibrating spring, pendulum and cantilever.
Figure 12.1 vibrating systems
Key
a is the amplitude.
A and C are the extreme positions of swing or oscillation (vibration).
B is the equilibrium or rest position.
Characteristics of oscillating systems
Oscillating systems have the following characteristics:
Amplitude (a) is the maximum displacement of an oscillating system from its resting position.
In Figure 12.1, amplitude is the distance between A and B or distance between B and C.
Amplitude is measured in metres (m) or centimeters (cm).
Displacement is the direction and distance from mean position. Displacement is measured in
metres (m) or centimeters (cm).
Period (T) is the time taken for one complete oscillation or cycle to be performed.
Period (T) is measured in seconds (s).
Frequency (f) is the number of complete oscillations or cycles produced in a unit time.
A cycle is a complete oscillation when an oscillating system moves from a starting point A to C
then back to A or moves from C to A then back to C.
Frequency is measured in hertz (Hz) or cycles per second
1 cycle per second = 1 hertz
Factors affecting frequency of an oscillating system
Simple pendulum.
Cantilever
A
B
C
a
C
B
A
a
a
a
248
1. For a pendulum
DISCUSSION
1. Calculate the frequency for each length of a string.
2. Plot a graph of frequency against length of a string.
3. From your results and the graph, what can you conclude?
4. Other variables are kept constant.
a. Explain what this means.
b. State two variables that can be kept constant.
Experiment 12.1
AIM: To find out whether the frequency of vibration of a pendulum depends on the length of the
string.
MATERIALS: A string, 50g bob, clamp stand, stop watch and ruler.
PROCEDURE:
1. Arrange the apparatus as shown below.
2. Pull the mass and release it to oscillate.
3. Record the time taken for 10 oscillations.
4. Record the results in Table12.1.
5. Repeat the experiment with length 20 cm, 30 cm, 40 cm and 50 cm.
Table 12.1
Length of a string
(cm)
Time for
10 oscillations(s)
Frequency = 10 oscillations
Time (s)
10
20
30
40
50
10 cm string
50 g mass
Figure 12.2
249
SUGGESTED RESULTS/EXPLANATIONS
1. Use the time found for 10 oscillations to calculate the frequency for each length of a
string using the formula given in the table.
2. After calculating frequencies, the graph will have a shape as shown in the sketch in
Figure 12.3.
3. Frequency decreases as the length increases and vice versa. As the length increases, the
string will take longer time to complete the 10 oscillations. Hence low frequency.
4. a. Other variables are not changed.
b. mass of the bob and type of the string
2. For a loaded spring
Frequency
(c/s)
Length (cm)
Figure 12.3
250
DISCUSSION
1. Plot a graph of frequency against mass.
2. Using the graph, determine how the mass affects the frequency.
3. For a cantilever
Experiment 12.2
AIM: To find out whether the mass affects the frequency of oscillation of a spring.
MATERIALS: Masses (50g, 100g, 150g and 200g), g-clamp, a stop watch and a spring.
PROCEDURE:
1. Set up the apparatus as shown in Figure 12.4:
2. Pull the mass downwards and leave it to vibrate freely.
3. Recording the time taken for 10 complete vibrations.
4. Calculate the frequency.
5. Repeat the experiment for the rest of masses (100g, 150g and 200g).
6. Record the results in Table 12.2 below:
Table 12.2
Mass (g)
Time for 10 vibrations (s)
Frequency = 10 vibrations
Time(s)
50
100
150
200
Figure 12.4
251
DISCUSSION
1. Plot a graph of frequency against mass.
2. Using the graph, determine how the mass affects the frequency.
Experiment 12.3
AIM: To find out whether the mass on the cantilever affects its frequency.
MATERIALS: Masses (50g, 100g, 150g and 200g), g-clamp, a stop watch and a
cantilever.
PROCEDURE:
1. Set up the apparatus as shown in Figure 12.5 below:
2. Pull the mass downwards and leave it to vibrate freely.
3. Recording the time taken for 10 complete vibrations.
4. Calculate the frequency.
5. Repeat the experiment for the rest of masses (100g, 150g and 200g) with a
fixed length, L.
6. Record the results in Table 12.3.
Table 12.3
Mass (g)
Time for 10 vibrations (s)
Frequency = 10 vibrations
Time(s)
50
100
150
200
Figure 12.5
252
DISCUSSION
3. Plot a graph of frequency against length.
4. Using the graph, determine how the length affects the frequency.
Exercise 12.1
In your groups, answer the following questions:
1. Define amplitude.
2. Explain why the amplitude of a vibrating spring decreases with increase in time.
3. Describe an experiment that you would carry out to find out whether the frequency of
Experiment 12.4
AIM: To find out whether the length of the cantilever affects its frequency.
MATERIALS: Mass, a metre rule (cantilever), g-clamp and a stop watch.
PROCEDURE:
1. Set up the apparatus as shown in Figure 12.6 below:
2. Pull the mass downwards and leave it to vibrate freely.
3. Recording the time taken for 10 complete vibrations.
4. Calculate the frequency.
5. Repeat the experiment by adjusting the length of the rule to 80 cm, 60 cm
and 40 cm with a fixed mass, M.
6. Record the results in Table 12.4 below:
Table 12.4
Length (cm)
Time for 10 vibrations (s)
Frequency = 10 vibrations
Time(s)
100
80
60
40
Figure 12.6
253
12.2 Waves
Oscillations or vibrations produce a wave. A wave is commonly taken as movement. We have
waves on the surface of the ocean or lakes and waves in the wind.
A wave is a means of disturbance or oscillation that travels through a medium or vacuum,
accompanied by a transfer of energy.
A wave motion is the transmission of energy from one place to another through a material or
vacuum.
RESULT
Experiment 12.5
AIM: To produce a wave.
MATERIALS: A tree and a rope.
PROCEDURE:
1. Tie a rope to the tree as shown in Figure 12.7.
2. Jerk the rope at the other end.
Figure 12.7
254
When you jerk a rope from the other end, humps and hollows are formed. These humps and
hollows form a wave as shown in Figure 12.8.
RESULT
When a pencil is moved up and down in the water at one end of the tray (ripple tank) or when a
vibrator touching the water is switched on ripples move away from the disturbance caused by the
pencil or vibrator. The ripples form a wave like the one shown in Figure 12.6.
Waves can be grouped into mechanical and electromagnetic waves.
Mechanical waves are the waves that require a medium for propagation. They cannot pass
through a vacuum. Examples of mechanical waves are sound wave and water wave.
Electromagnetic waves are the waves that do not need a medium for propagation.
Electromagnetic waves can pass through a vacuum. Examples of electromagnetic waves are
radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, x-rays and
gamma rays.
Figure 12.8 A wave
Experiment 12.6
AIM: To investigate water waves.
MATERIALS: Transparent tray (ripple tank), water and pencil (or vibrator).
PROCEDURE:
1. Fill a shallow tray (ripple tank) with water.
2. Move a pencil up and down in the water at one end of the tank or let a vibrator
just touch the surface of the water and switch it on. Record the observations.
255
Characteristics of a wave
A wave has the following characteristics.
1. Amplitude (a)
Wave amplitude is the maximum displacement of a particle from its resting position.
It is measured in centimetres (cm) or metres (m).
From Figure 12.9 the amplitude of the wave is 10 cm.
2. Frequency (f)
Wave frequency is the number of complete oscillations or cycles produced per second.
Frequency (f) = number of complete cycles (c)
Time in seconds (s)
Frequency is measured in hertz (Hz) or c/s.
1 c/s = 1 Hz
For example: If the rope makes 10 complete cycles in 5 seconds,
Frequency = 10 cycles
5 seconds
Frequency (f) = 2Hz
Figure 12.9 Showing amplitude of a wave
256
3. Period (T)
Wave period is the time taken for one complete oscillation or cycle to be performed.
Period (T) = Time taken
Number of complete cycles
From the equation shown above, period (T) is the inverse of frequency (f).
Therefore,
T = 1
f
Hence f = 1
T
Worked example
Figure 12.10 is a diagram showing a wave motion.
a. Calculate the amplitude of the wave.
b. Calculate the frequency of the wave.
c. What is the period of the wave?
Solution
a. Amplitude = 1 m
Figure 12.10
Displacement
(m)
Time
(s)
4
257
b. Frequency = Number of complete cycles
Time taken
Frequency = 1 cycle
4 seconds
Frequency = 0.25 Hz
c. Period (T) = 1
f
T = 1
0.25 Hz
T = 4 seconds
4. Wavelength (λ)
Wavelength is the distance between two successive particles which are at the same point after a
complete oscillation in their paths and are moving in the same direction. These are the distances
occupied by one complete oscillation.
Wavelength is represented by a Greek symbol called Lambda (
).
Wavelength is measured in metres (m).
The wavelength of a wave can be found by the following methods:
a. Checking the distance covered by one complete cycle.
Wavelength = distance covered by one complete cycle
Figure 12.11 showing the wave length of the wave
258
In Figure 12.11, the distance covered by one complete cycle is 2 m. Therefore, the wavelength
of a wave is 2 m.
b. Wavelength = Total distance covered by the wave
Number of cycles
Worked example
Calculate the wavelength of a wave in Figure 12.12.
Solution
Wavelength = Total distance covered
Number of cycles
Wavelength = 10 m
2 cycles
Wavelength = 5 m.
5. Velocity
Wave velocity (speed) is the distance covered by the wave in a unit time.
Velocity = distance covered by the wave
Time taken
Worked example
A water wave covered a distance of 50 m in 10 seconds. Calculate its velocity.
Solution
d = 50 m t = 10 s
Figure 12.12
Distance (m)
10 m
259
v = d
t
v = 50 m
10 s
v = 5 m/s
6. Phase
Wave phase is the orientation of wave pulses in space with respect to the origin of the
wave.
Wave front
Figure 12.13(a) the waves are in phase
Figure 12.13 (b) the waves are out of phase
260
A wave front is any line or section taken through an advancing wave which joins all points
which are in the same position in their oscillations. Wave fronts are usually at right angles to the
direction of the waves and can have any shape, e.g. circular and straight wave fronts.
Circular wave front can be produced by dropping a spherical object in water. The spherical
object causes disturbance in water. Water forms circular patterns called circular wave front.
Straight wave front can be produced by dipping a straight edge in water. The straight edge causes
disturbance in water. Water produces straight wave fronts.
Figure 12.14 circular wave front
Source of disturbance
Wave front
Figure 12.15 straight wave front
Wave front
Direction of the wave
Source of disturbance
261
Exercise 12.2
In your groups, answer the following questions:
1. Figure 12.16 shows a wave.
a. On the diagram, label the following:
i. Amplitude (a) ii. Wavelength (λ)
b. Calculate the wavelength of the wave.
2. The frequency of a wave is 200 Hz. Calculate the period (T) of a wave.
3. A straight wave front is produced by a vibrator in a ripple tank.
a. What is the wavelength of the ripples if there are 10 complete waves in a
distance of 50 cm?
b. Calculate the frequency of the ripples if 10 complete waves were produced in a
minute.
Figure 12.16
Figure 12.13
262
12.3 Types of waves
The two basic types of waves are longitudinal wave and transverse wave.
Longitudinal waves
When the end of a spring is moved backwards and forwards, the sections of the coil are pulled
together and released. These sections are known as compressions and rarefactions. This
produces a travelling wave effect. In this wave oscillations are backwards and forwards. The
wave is called longitudinal wave.
A longitudinal wave is a wave in which the direction of the vibrating particles (oscillations) is
the same as the direction of a wave itself OR it is a wave in which the displacements are parallel
to the direction of a wave itself.
Compressions: These are regions of high pressure and density along a longitudinal wave where
particles are squeezed.
A longitudinal wave can be demonstrated by using a spring as shown in
Figure 12.17.
Rarefaction
Figure 12.17 demonstrating longitudinal wave in a spring
263
Rarefactions: These are regions of low pressure and density along a longitudinal wave where
particles are spaced.
An example of a longitudinal wave is a sound wave.
Transverse waves
A transverse wave can be demonstrated by a spring as shown in Figure 12.19.
When a string is moved up and down in the direction perpendicular to its length, the particles of
the rope near the end pull the next particles sideways then they pull the next particles and so on.
Figure 12.18 longitudinal wave
Figure 12.19 demonstrating transverse wave by using a spring
264
Sideways movements are passed from turn to turn and a traveling wave effect is produced. In
this way, there is a transferring of energy from one end of the string to the other.
When the oscillations are up and down or from side to side as shown in Figure 12.19, the wave
produced is called Transverse Wave.
In a transverse wave the oscillations are perpendicular (at right angles) to the direction of the
wave itself.
A transverse wave is a wave in which direction of vibrations or oscillations is perpendicular (at
right angle) to the direction of propagation of the wave.
Crests or peaks: These are points where a wave causes maximum positive displacement of the
medium.
Troughs: These are points where a wave causes maximum negative displacement of the
medium.
Examples of transverse waves are water wave and light wave.
NOTE: In a longitudinal wave the wavelength can be defined as the distance between two
successive compressions or the distance between two successive rarefactions.
In a transverse wave the wavelength can be defined as the distance between two successive
crests or the distance between two successive troughs.
Figure 12.20 a transverse wave
265
12.4 Wave properties
Waves have the following properties:
a. Reflection
Exercise 12.3
In your groups, answer the following questions:
1. Define
a. Compression b. Rarefaction c. Crest d. Trough
2. With the aid of well labeled diagrams, discuss the differences between transverse
and longitudinal waves.
3. Explain how you can demonstrate transverse and longitudinal waves.
Experiment 12.7
AIM: To investigate wave reflection.
MATERIALS: Ripple tank, water and vibrator.
PROCEDURE:
1. Fill a shallow ripple tank with water.
2. Let a vibrator just touch the surface of the water and switch it on to create waves
moving down the tank.
3. Place a horizontal metal strip at an angle to the direction of the wave. Record what
happens to the wave.
266
When an obstacle is placed in the path of the wave it changes its direction. The wave is bounced
off. This effect is called reflection.
Reflection is defined as the bouncing off of waves when an obstacle is placed in their path.
The laws of reflection
i. The angle of incidence is equal to the angle of reflection.
This means that the wave leaves the surface at the same angle it arrives.
Angle of incidence (i) = angle of reflection (r).
Figure 12.21 Reflection in a water wave
Figure 12.22 Reflection in a light wave
267
ii. The incident ray, the normal and the reflected ray all lie in the same plane.
This means that all three could be drawn on the same flat piece of paper.
b. Refraction
In Experiment 12.8, water waves are made to travel from a deeper region to a shallow region. In
this case, the following happens:
i. The wavelength decreases. In a deep region, a wavelength is greater because of high speed
while in a shallow region the wavelength is shorter because of the decrease in speed.
ii. The wave appears to change direction. The wave changes direction because it changes
speed when traveling from a deep region to a shallow region. In a deep region, the wave
travels faster and it slows down when it enters the shallow region. This apparent bending
of the wave is called refraction.
In both regions, the frequency remains the same.
Refraction is the bending of a wave when it changes its speed or velocity.
Water waves undergo refraction or bending when they enter shallow water.
In shallow water the water waves are slowed down. In shallow water, the wave length of a water
wave is reduced.
Experiment 12.8
AIM: To investigate wave refraction.
MATERIALS: Ripple tank, water, vibrator and glass or plastic sheet.
PROCEDURE:
1. Set up a ripple tank with a piece of material that will create a shallow section in
the tank.
2. Let a vibrator just touch the surface of the water in a deep region and switch it on
to create waves that will travel down the tank from “deeper” end and across the
shallow section.
3. Record the observations.
268
A light wave undergoes refraction when it moves into a medium of different density which
causes it to travel at a different speed or velocity.
When a ray of light travels from a less dense medium (e.g. air) to a denser medium (e.g. glass) it
bends towards the normal because it travels with less speed in the denser medium.
When the ray of light travels from a denser medium (e.g. glass) to a less dense medium (e.g. air)
it bends away from the normal because it travels with greater speed in the less dense medium.
Figure 12.23 Refraction of a water wave.
Figure 12.24 Refraction of a light wave
i = angle of incidence r = angle of reflection
269
Laws of refraction
i. The incident and refracted rays are on opposite sides of the normal at the point of
incidence.
ii. When it comes to the incident ray, the normal and the refracted ray lie in the same plane.
iii. The value of angle of the ratio of the angle of incidence to the angle of refraction is the
same for light passing from one given medium into another. For example, dividing sine i
by sin r, when the ray of light passes from air into glass, always produces the same number
whatever the angle of incidence is.
= constant
This constant is called refractive index. Refractive index, when referring to light, is the
light-bending ability of a medium. For example, light bends more in glass than in water. This is
also known as the optical density.
= refractive index
Table 12.4 shows the refractive index of some media.
Table 12.4 Refractive index of some media
Medium
Refractive index
Water
1.33
Paraffin
1.44
Perspex
1.49
Glass
1.52
Diamond
2.42
Refractive index can also be calculated by dividing the speed of light in a vacuum or in air by the
speed of light in a medium.
refractive index =
Worked examples
1. Calculate the refractive index if:
a. the sin i = 65
0
and sin r = 40
0
b. the speed of light in air is 3.0 x 10
8
m/s and its speed when it enters the water is
2.25 x 10
8
m/s.
sin i
sin r
sin i
sin r
speed of light in a vacuum (air)
speed of light in a medium
270
Solution
a. Refractive index =
Refractive index =
Refractive index =
Refractive index =1.4
b. Refractive index =
Refractive index =
Refractive index = 1.33
2. The ray of light from air forms an angle of incidence of 80
0
at the surface of the glass. If
the refractive index of glass is 1.5, calculate the angle of refraction.
Solution
Refractive index =
1.5 =
Sin r =
Sin r = 0.6565
Angle of refraction, r = sin
-1
0.6565
Angle of refraction, r = 41.0
0
sin i
sin r
sin 65
0
sin 40
0
0.906
0.643
speed of light in air
speed of light in water
3.0 x 10
8
2.25 x 10
8
sin i
sin r
sin 80
0
sin r
sin 80
0
1.5
271
c. Diffraction
Diffraction is the spreading out of waves when passing through a slit or a gap of an obstacle.
Diffraction in a narrow gap
When waves pass through a narrow gap, there is more or stronger diffraction (spreading out)
because the waves pass with greater pressure.
Diffraction in a wide gap
When waves pass through a wide gap, there is less or weak diffraction (spreading out) because
the waves pass with less pressure.
Figure 12.25 diffraction of a wave in a narrow gap
Figure 12.26 diffraction of a wave in a wide gap
272
Diffraction in two slits or gaps
Figure 12.27 shows the water and light waves approaching two slits, S1 and S2 in an obstacle.
When a wave is diffracted in two slits, diffracted waves overlap and cause interference.
d. Interference
Interference is caused if two identical sets of waves travelling through the same region of water
result in either reinforcing or cancelling each other.
Types of interference
i. Constructive Interference
Constructive interference is caused if two identical waves are in phase, both are moving in the
same direction. The crest of one wave meets with the crest of another wave while the trough of
one wave meets with a trough of another wave. These waves are always in phase, meaning they
have a phase difference of 0
0
. During constructive interference, the amplitude of the resultant
wave is doubled.
Figure 12.27 Diffraction of a wave in two slits
273
ii. Destructive interference
Destructive interference is caused when two identical waves move in opposite directions. When
the waves meet, the crest of one wave coincides with the trough of the other wave, while the
trough of the other wave coincides with the crest of the other wave. The waves are out of phase
by 180
0
.
Figure 12.28 constructive interference
274
This results in no wave or no movement. Hence the property is also called Cancellation.
12.5 Wave equation
Figure 12.29 destructive interference
Exercise 12.4
In your groups, answer the following questions:
1. a. Define reflection.
b. State two laws of reflection.
2. Explain why the water wave changes direction when it moves from a deep region to
a shallow region.
3. With the aid of a well labeled diagram, show the path of a light wave from air, within
and beyond the glass.
4. Refractive index of water is 1.4. Light wave enters water at an angle of incidence of
25
0
. Calculate the angle of refraction.
5. Explain the difference between waves passing through narrow and wide gaps.
6. Describe:
a. constructive interference.
b. destructive interference.
275
The speed or velocity, frequency and wavelength are linked by an equation called wave
equation.
If the speed of a wave is V in m/s, the frequency is f in Hz and the wavelength is λ in m.
The wave equation becomes:
V = f x
To derive the wave equation:
Velocity of a complete cycle = distance covered by a complete cycle
Time taken by a complete cycle to be performed
Distance covered by a complete cycle(d) = wavelength (λ)
Time taken for a complete cycle to be performed = Period (T)
Velocity of a complete cycle: V = λ
T
Which can also be written as: V = 1 x λ
T
But 1 = f
T
Therefore, the wave equation becomes: V = f x λ
Worked examples
1. The wave crests in a ripple tank are 3 mm apart. Calculate the speed of the wave if the
frequency of the vibrator is 15 Hz.
Solution
λ = 3 mm = 0.003 m f = 15 Hz
V = f x λ
V = 15 Hz x 0.003 m
V = 0.045 m/s
2. Calculate the frequency of a wave of wavelength 100 m and its velocity is 3 x 10
8
m/s.
Solution
V = f x
f = V
276
f = 3 x 10
8
m/s
100
f = 3 x 10
6
Hz
3. Nyasa broadcasting station broadcasts a radio wave of speed 3 x 10
8
m/s on a frequency
of 750 kHz. Find the wavelength of this wave
Solution
V = 3 x 10
8
m/s
f = 750 kHz = 7.5 x 10
5
Hz
= ?
V = f
= V
f
= 3 x 10
8
m/s
7.5 x 10
5
Hz
= 400 m
Summary
Oscillations or vibrations are complete upward or downward movements of an object about its
fixed position (rest position or equilibrium position). Oscillations or vibrations can also be
defined as complete to and fro movements of an object about its fixed position (rest position or
equilibrium position).
Exercise 12.5
In your groups, answer the following questions:
1. A turning fork produces 200 cycles in 4 seconds. Find the wavelength of this sound
wave if the speed of sound in air is 330 m/s.
2. A radio wave is transmitted at a frequency of 10 kHz. If the speed of a radio wave is 3
x 10
8
m/s, calculate
a. its wavelength
b. the time taken for the wave to travel a distance of 5 km.
277
Characteristics of oscillating systems are:
Amplitude: the maximum displacement of an oscillating system from its resting position.
Displacement is the direction and distance from mean position. Displacement is measured in
metres (m) or centimeters (cm).
Period: the time taken for one complete oscillation or cycle to be performed.
Frequency: the number of complete oscillations or cycles produced in a unit time.
Factors that affect frequency of an oscillating system are
Pendulum: length of the string.
Loaded spring: mass of the bob.
Cantilever: mass at its end.
A wave is a means of disturbance or oscillation that travels through a medium or vacuum,
accompanied by a transfer of energy.
A wave motion is the transmission of energy from one place to another through a material or
vacuum.
Characteristics of a wave are amplitude, frequency, velocity, period, phase and wavelength.
The two types of waves are transverse and longitudinal waves.
The properties of waves are reflection, refraction, diffraction and interference.
The equation which relates speed (velocity) of a wave, its frequency and wave length is called
wave equation.
V = f x
Student assessment
1. Define the following:
a. Amplitude
b. Frequency
c. Refraction
d. Wave front
278
2. Figure 12.30 is a diagram showing a wave motion
a. What type of wave is represented in the diagram?
b. What is the amplitude of the wave?
c. What is the wavelength of the wave?
d. If the period of the wave is 10 seconds, calculate its frequency.
e. Calculate the velocity of the wave.
3. Draw a diagram of a transverse wave. In the diagram, indicate amplitude, wavelength and
direction of the wave motion.
4. Figure 12.31 shows a wave form of a pendulum.
a. What is the type of the wave shown in the Figure 12.31?
b. Work out the frequency of the wave.
c. Calculate the period of the wave.
d. State the energy changes that take place in a pendulum.
e. What causes the amplitude of oscillation of a pendulum decrease as time increases?
5. A tuning fork produces 500 cycles in 5 seconds. Find the wavelength of this sound wave
if the speed of sound in air is 340 m/s.
Figure 12.31
Figure 12.30
279
6. The frequency of a microwave is 100 kHz; calculate its wavelength given the speed of
electromagnetic waves in a vacuum or air = 3 x 10
8
m/s.
7. State two properties of the water waves.
8. What can be said about the phase difference of two identical sets of waves if they cause:
a. Constructive interference
b. Destructive interference.
9. With the aid of well labeled diagrams, explain the formation of constructive interference
and destructive interference.
10. With the aid of well labeled diagrams, explain the difference between longitudinal and
transverse waves.
11. Figure 12.32 shows a ray of light incident on air-glass boundary.
Complete the path of the ray of light to show
a. reflection
b. refraction
12. Given the spring, bobs (50 g, 100 g, 150 g and 200 g), clamp, stopwatch and ruler,
describe an experiment that you would carry out to find out whether the frequency of a
vibrating spring depends on the mass of the bob.
13. Explain why the amplitude of a vibrating cantilever decreases with increase in time.
Glass
Air
Figure 12.32
280
14. a. Define
i. transverse wave
ii. longitudinal wave
b. Give an example of:
i. a longitudinal wave.
ii. a transverse wave.
15. A transverse wave travels a distance of 50 cm in 5 seconds and its wavelength is 2 m.
Calculate:
a. its velocity.
b. the frequency of the wave.
16. a. Define oscillations in a pendulum.
b. In a swinging pendulum, explain what happens to the following as the time increases:
i. Amplitude
ii. Frequency.
17. A form three student wanted to find out if the frequency of a vibrating cantilever is
affected by mass. Describe an experiment that she would carry out.
18. State two factors that affect the frequency of a pendulum.
19. Describe an experiment that you would carry out to show that the frequency of a
vibrating pendulum depends on the length of a string.
281
Sound
Objectives
At the end of chapter 13, you must be able to:
Describe experimentations to show that sound is
produced by vibrating bodies
Discuss free vibrations, forced vibrations, natural
frequency and resonance
Explain the nature of sound waves
Explain the factors affecting the speed of sound
13.1 Production of sound
Sound is a wave which belongs to a type of a wave called longitudinal wave. The direction of the
particles is the same as the direction of the wave itself.
Production of sound
Sound waves are produced by vibrations of the vibrating systems. Examples of objects that can
produce sound are loudspeaker, tuning fork, toothed wheel, siren etc.
Chapter 13
Experiment 13.1
AIM: To show that sound is produced by vibrating objects.
MATERIALS: Person, elastic band and tuning fork.
PROCEDURE:
1. Put a finger on the throat of a person who is speaking. What do you observe?
2. Stretch an elastic band and pluck it. What do you observe?
3. Tap a tuning fork. What do you observe?
282
RESULTS/EXPLANATIONS
1. When you put a finger on the throat of someone who is speaking, the person starts
humming. The humming can give a clue to how sound is produced.
2. If an elastic band is stretched and plucked, it will be seen vibrating and it will produce a
humming sound. In this case, the elastic band represents the tissue called the vocal chords
found in the throat which vibrate to produce sound.
3. When a tuning fork is tapped gently it will vibrate. Sound is produced as it vibrates.
CONCLUSION
From the above observations, it shows that sound is produced by vibrations caused by vibrating
objects.
Amplitude and loudness of sound
When a loudspeaker cone vibrates, it moves forwards and backwards. The maximum distance
the loudspeaker cone moves forwards and backwards is called amplitude. The amplitude of a
sound wave is the maximum distance the vibrating system moves backwards and forwards from
its rest position.
The amplitude of a sound wave produced increases with an increase in the amplitude of a
loudspeaker cone. An increase in amplitude causes more sound energy to travel out through the
air every second to the ear. Hence the sound becomes louder.
Therefore, the loudness of sound depends upon the amplitude of the wave that produces it.
Figure 13.1 effect of amplitude of the wave on loudness of sound
283
Frequency and pitch of sound
Sound waves are created by vibrations or oscillations. The number of oscillations per second is
called frequency. Frequency of a sound wave can also be considered as the number of
wavelengths the wave can produce per second.
Sound waves of different frequencies sound different to the ear. Sound wave of high frequency is
heard as a note said to be of high pitch. Sound wave of low frequency is heard as a note said to
be of low pitch.
Sound of high frequency has a note of high pitch and a short wavelength. Sound of low
frequency has a note of low pitch and long wavelength.
Audible sound
Audible sound range is the sound of the frequency which can be detected by human ear. The
frequency of audible sound ranges from 20 Hz to 20 kHz (20 000 Hz).
Figure 13.2 effect of frequency on the pitch of sound
284
Ultrasonic sound
Ultrasonic sound range is the sound of frequency that cannot be detected by human ear but other
animals e.g. dogs, bats and fish.
This is a sound with very high frequency of greater than 20 kHz (20 000Hz).
Table 13.1 frequency of vibrations and pitch notes
Pitch
Frequency (Hz)
High
Low
Upper limit of hearing
Whistle
High note (soprano)
Low note (bass)
Drum note
20 000
10 000
1000
100
20
285
Exercise 13.1
In your groups, answer the following questions:
1. Table 13.2 shows the frequencies of sound waves produced by five loud speakers.
Table 13.2
Loud speaker
Frequency (Hz)
1
300
2
200
3
400
4
100
5
250
Which sound has:
a. the highest frequency? b. the lowest frequency?
c. the highest pitch? d. lowest pitch?
e. the longest wavelength? f. the shortest wavelength?
2. Explain why high amplitude produces a louder sound.
3. Figure 13.3 below shows sound waves produced by sirens A and B.
Figure 13.3
Which sound wave;
a. is louder? b. has a higher pitch?
Siren A
Siren B
286
13.2 Free and forced vibrations
Free vibrations
When the string in Figure 13.4 is plucked, it vibrates freely. It continues vibrating when left
alone. This vibration is called free vibration. The frequency with which it vibrates is called
natural frequency.
Examples of natural frequency are a child on a swing who has been pushed once only, a punch-
ball which has received just one punch, a simple pendulum and a tuning fork which has been
struck.
Forced vibration
When the string in Figure 13.4 is plucked continually its frequency is determined by the person
plucking it, when the child on the swing is pushed continually its frequency is determined by the
person pushing the swing. Similarly, when a boxer punches the punch-ball continually, the
frequency with which the punch-ball oscillates is determined by the frequency with which the
boxer punches. These objects are not vibrating with their natural frequency but they are forced to
vibrate. These vibrations are called forced vibrations. The frequency with which these objects
vibrate is called forced frequency.
When an object is vibrated continually it reaches its extreme position. Therefore, forced
vibrations become much larger in amplitude.
Figure 13.4 free vibrations
String
Mass
287
Resonance
If you strike a tuning fork once and leave it to vibrate, it produces natural frequency due to
natural vibration. When the vibrating tuning fork is brought closer to the air column, air column
is forced to vibrate at the same frequency as the tuning fork.
Therefore, the air column produces forced frequency.
When the natural frequency of the natural vibrating tuning fork equals the forced frequency of
the forced vibrating air column, the resonant point is reached. This phenomenon is called
Resonance.
Resonance takes place when a body is made to vibrate at its natural frequency by vibrations
received from another vibrating source of the same frequency.
Resonance is a phenomenon (happening) that needs two vibrations:
a. Forced vibration
b. Natural vibration
Resonance takes place when natural frequency equals forced frequency.
“Forced vibration” frequency = “natural vibration” frequency
Demonstrating resonance by using Barton’s pendulum
Figure 13.5 Barton’s pendulum
288
Every object has its own natural frequency of vibration. The object will vibrate with that
frequency when it has the opportunity.
Figure 13.5 shows a stretched string to which ten pendulums of different lengths are attached.
When the metal ball is set swinging, it forces all the ten bobs swinging as followers. The bob
with the same length of string as the metal ball will swing with a much larger amplitude. A’s
frequency equals the metal ball’s frequency because they have the same length. Therefore, A
resonates with the metal ball.
Experiment 13.1
AIM: To investigate resonance.
MATERIALS: Resonance tube, clamp stands, tuning forks (frequencies 400 Hz,
500 Hz and 600 Hz) and water.
PROCEDURE:
1. Set up an experiment as shown in Figure 13.6.
2. Vibrate a tuning fork of a frequency 400 Hz.
3. Take the vibrating tuning fork and hold it over the mouth of the
tube with water.
Figure13.6
289
EXPLANATION/CONCLUSION
When the vibrating tuning fork is brought on the mouth of the tube with no air column you will
not hear any sound. As the level of water is decreased and the air column is increased you will
hear the sound increase. Increasing the length of air column further gives the maximum sound.
The loud sound is heard when the air column reaches a certain critical length. This is called the
position of resonance.
The same experiment can be carried out by using a glass tube placed in a jar with water. The
length of the air column is varied by raising or lowering the tube.
Figure 13.7 Resonance in a jar with water
4. Vary the depth of the water in the tube until the air column is
made to resonate with the turning fork. Record the length of the
air column (l) that was required to produce resonance.
5. Repeat the experiment with other tuning forks of frequency 500
Hz and 600 Hz. Record the air column used to produce resonance
in each case.
6. Record your results in Table 13.3.
Table 13.3
Frequency (f)
Length of air column (l)
400 Hz
500 Hz
600 Hz
290
Figure 13.8 shows the graph of how the amplitude of sound varies with frequency in order to
reach the position of resonance.
Fb = Resonant frequency
B = Resonant point because there is maximum amplitude of vibration
Uses of resonance
Resonance is used in the following;
a. Child’s swing
b. Driving a car
c. Swinging bridge
d. A diver jumps up and down at the board’s natural frequency when he wants to perform a
very high dive. This resonance increases the amplitude of the springboard and the diver
has no difficulty in reaching the required height.
Resonance can be a nuisance and dangerous. Resonance can cause breaking in swinging bridges
when people are marching on it. When the natural frequency and forced frequency are equal the
bridge vibrates violently.
Figure 13.8 resonance graph
291
13.3 Nature of sound waves
Propagation of sound
Sound waves are caused by vibrations.
Propagation (spreading) of sound is the way by which sound travels from side to side or from
where it is produced to where it is heard.
When the bell is struck, it vibrates. The vibrations compress then stretch the air particles, as
shown in Figure 13.9. The compressed (squeezed) air particles form a region of high pressure
called compression(C). The stretched (spaced) air particles form a region of low pressure called
rarefaction(R). The compressions and rarefactions travel forward to the ear. Air transmits a
longitudinal wave. The wave is known as sound wave.
Exercise 13.2
In your groups, answer the following questions:
1. With the aid of a diagram, explain how you can demonstrate free and forced
vibrations.
2. What do you understand by the word ‘resonance’?
3. Describe with the aid of a diagram, an experiment you would carry out to obtain
resonance between a tuning fork and a column of air in a bottle.
4. State two situations where resonance is good.
5. State one instance where resonance is dangerous.
292
Figure 13.10 shows how air pressure varies along the path of a sound wave.
Transmission of sound in air and in a vacuum
Figure 13.9, shows that a sound wave is transmitted in air as a longitudinal wave.
A sound wave needs a medium or material to travel through because it is a longitudinal wave and
requires a material that can pass on oscillations. Sound cannot travel in a vacuum.
Sound waves can travel through air, solids and liquids.
Figure 13.9 transmission of sound in air
Figure 13.10 air pressure along the path of a sound wave
293
RESULTS/EXPLANATIONS
Before starting the vacuum pump: When the glass jar is well closed and the switch is closed,
the hammer hits the gong. The sound of an electric bell is heard.
After starting the vacuum pump: When the vacuum pump is started, the sound of an electric
bell becomes fainter until it cannot be heard. But the hammer can be seen striking the gong.
When the air is pumped in once more, the sound of an electric bell is heard once again.
CONCLUSION
Therefore, sound waves can only be heard if there is a material or medium present to pass
oscillations, so it is not possible for sound to travel through a vacuum.
Experiment 13.2
AIM: To demonstrate that sound waves do not pass through a vacuum.
MATERIALS: Electric bell, power supply, switch, connecting wires, bell jar, cork and
vacuum pump.
PROCEDURE:
1. Suspend an electric bell inside a bell jar.
2. Connect a vacuum pump to the bell jar as shown in Figure 13.11.
3. Close the switch before starting the vacuum pump.
Explain your observations.
4. Start the vacuum pump.
Explain your observations.
Figure 13.11
294
Speed of sound
Speed is the distance covered per unit time. The speed of sound is mainly measured by using the
reflection of sound.
Reflection of sound
Reflection of sound takes place when it strikes an obstacle, e.g. a wall or a cliff.
The reflected or bounced off sound wave is called an echo.
An echo can be used to measure the speed of sound in air and measure the depth of a sea.
Measuring the speed of sound using an echo
Experiment 13.3
AIM: To determine the speed of sound in air
MATERIALS: Stopwatch, toy gun, measuring tape and a high wall.
PROCEDURE:
1. Measure a distance of 100 m from the high wall by using a measuring tape.
2. Stand at a distance of 100 m (point A) from the high wall (point B).
3. Fire a toy gun at point A. Start the stopwatch as soon as you fire a toy gun.
Stop the stop watch as soon as you hear an echo.
Record the time taken from the time the gun was fired to the time an echo was
heard.
High wall
100m
A B
Figure 13.12
295
RESULTS/EXPLANATIONS
The time taken from when the sound wave was produced to when an echo was heard is t. The
time (t) is for the distance A to B then back to A.
The distance covered is 2d.
Therefore, the speed of sound using an echo can be estimated as follows:
Speed = distance covered by the sound wave
time taken
Speed of sound = 2d
t
S = 2d
t
Note that the speed of sound without using an echo is:
S = d
t
Transmission of sound in different media
Sound is transmitted at different speeds in different media.
The speed of sound varies considerably depending on the material through which the waves are
traveling.
The following are the speeds of sound in different media or materials:
Air = 330 m/s (dry air, at 0
0
C)
Water = 1400 m/s (at 0
0
C)
Solid = 500 m/s
Worked examples
1. An observer sees a flash of a gun being fired and hears the sound 2.4 seconds later. If the
distance from the gun to the observer is 816 m/s, calculate the speed of the sound in air.
Solution
In this situation, there is no echo being produced.
Therefore, s = d
t
d = 816 m t = 2.4 s
296
s = d
t
s = 816 m
2.4 s
s = 340 m/s
2. A girl standing in front of a hill produces a sound and hears an echo 4 seconds later. If
the speed of sound in air is 330 m/s, how far is the hill from the girl?
Solution
s = 330 m/s
t = 4 s
d =?
s = 2d (since there is an echo)
t
d = s x t
2
d = 330 m/s x 4s
2
d = 660 m
3. A boy standing between two cliffs is 850 m from the nearest cliff. When he fires a gun,
the first echo is heard after 5 seconds, and the second echo is heard 8 seconds later.
Calculate:
a. The speed of sound.
b. The distance between the cliffs.
Solution
a. To find the speed of sound in air
s = 2d
t
d = 660 m
t = 5 s
s =?
s = 850 m x 2
5 s
s = 340 m/s
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b. The distance between two cliffs = distance from the boy to the nearest cliff
+ distance from the boy to the furthest cliff
Distance from the boy to the nearest cliff = 850 m
Distance from the boy to the furthest cliff can be found as follows:
s = 2d
t
d =?
s = 340 m/s
t = 8 s
d = s x t
2
d = s x t
2
AC = 340 m/s x 8 s
2
d = 1360 m
Distance between two cliffs = 850 m + 1360 m
d = 2210 m
Wave equation
The speed of sound can also be found by using the wave equation:
V = f x λ
Whereby V is velocity (speed) in m/s, f is frequency in Hz and λ is wavelength in m.
Worked examples
1. Find the speed of the sound wave if its wavelength is 3.4 m and its frequency is 100 Hz.
Solution
λ = 3.4 m f = 100 Hz
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V = f x λ
V = 100 Hz x 3.4 m
V = 340 m/s
2. A tuning fork produces 250 cycles in 2.5 seconds. Find the wavelength of this sound if the
speed of sound in the air is 340 m/s.
Solution
f = 250 cycles
2.5 s
f= 100 Hz
V= 340 m/s
λ =?
V= f x λ
λ = V
f
λ = 340 m/s
100 Hz
λ = 3.4 m
Exercise 13.3
In your groups, answer the following questions:
1. Yankho sees steam start to come from a factory whistle and she hears the sound
5 seconds later. If the speed of sound in the air is 340 m/s, how far is she from the
whistle?
2. How long will a sound wave travel 1000 m if the speed of sound in the air is 330
m/s?
3. The loudspeaker placed 80 m in front of a large vertical wall produces sound
waves. If the speed of sound in the air is 340 m/s, calculate the time taken for the
sound wave to be reflected back to the loudspeaker.
4. A man standing between two cliffs claps two blocks of wood. He hears two
echoes, the first after 2 seconds and the next after 3 seconds. The speed of sound in
air is 340 m/s. What is the distance between the cliffs?
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13.4 Factors that affect the speed of sound
The following are the factors that affect the speed of sound in a media:
a. Direction of wind: In air, a sound wave travels faster when it is traveling in the direction
of the wind and vice versa.
b. Temperature: In air, the speed of sound can increase as the temperature increases
without altering the pressure. The speed of sound increases with an increase in
temperature when pressure is constant because the air expands and becomes less dense.
Therefore, the compressions and rarefactions can easily be transmitted.
For example, the speed of sound in the air at 0
0
C is 330 m/s while its speed at 25
0
C is 340
m/s.
c. Strength of intermolecular forces in a medium
The sound is slowest in gases, more rapidly in liquids and fastest in solids because the
forces become stronger. Stronger forces make particles to be tightly packed. The
oscillations are passed on more rapidly in a medium with tightly packed particles.
Summary
Sound is produced by vibrating bodies e.g loudspeaker, tuning fork, toothed wheel, siren, bell,
guitar, hacksaw blades, ruler, empty bottle whistle and drum.
Amplitude of the vibrating system affects the loudness of the sound produced. The higher the
amplitude, the louder the sound and vice versa.
The frequency of a sound wave affects the pitch of sound. Sound of high frequency has a note of
high pitch and vice versa.
Natural vibration is a vibration of any object when it is set to oscillate. A frequency at which
any object vibrates freely is called natural frequency.
Forced vibration is the vibration of an object when it is forced to vibrate at a frequency other
than its natural frequency. A frequency at which an object is forced to vibrate is called forced
frequency.
Resonance takes place when a body is made to vibrate at its natural frequency by vibrations
received from another vibrating source of the same frequency.
Resonance takes place when natural frequency equals forced frequency.
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A sound wave is transmitted in a medium by means of compressions and rarefactions. A Sound
wave is a longitudinal wave. Sound requires a medium for propagation.
Speed of sound can be determined experimentally by using an echo.
The formula for the speed of sound is:
Speed = distance covered
time taken
When there is an echo, speed is found as:
Speed = 2 x distance covered
time taken
The speed of sound is different in different media as shown below:
Air = 330m/s
Liquid = 1400 m/s
Solid = 5000 m/s
Factors that affect the speed of sound in media are direction of the wind (in the air), temperature
and strength of the bonds.
Student assessment
1. Explain what is meant by
a. a compression
b. a rarefaction
c. a longitudinal wave
2. Figure 13.13 shows how the air pressure varies along the path of a sound wave.
Figure 13.13
Air
pressure
Normal
pressure
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a. On Figure 13.13, mark the points
i. that have high air pressure with letter H.
ii. that have low air pressure with letter L.
iii. that represent compressions with letter C.
iv. that represent rarefactions with letter R.
b. Describe the motion of air particles along the path of a sound wave.
c. Calculate the velocity of the wave if its wavelength is 2.2 m and it has a frequency of 150
Hz.
3. a. Define an echo.
b. Explain how an echo can be used to find the depth of the sea.
4. Calculate the wavelength of a sound wave that is produced by a source vibrating with a
frequency of 50 Hz. The speed of sound in the air is 340 m/s.
5. Explain the difference between audible sound and ultrasonic sound.
6. Explain why sound does not travel through a vacuum.
7. A boat hears the echo from a sound wave 5 seconds after it has been emitted. If the speed
of sound in the water is 1400 m/s, calculate the depth of the sea.
8. Describe an experiment that you would carry out to show that sound wave requires a
medium for propagation.
9. Figure 13.14 shows sound waves shown on the oscilloscope.
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Figure 13.14
Which sound
a. is the loudest?
b. has the highest frequency?
c. has the highest amplitude?
d. has the lowest pitch?
e. will sound the lowest?
10. A girl standing 400 m away from the cliff shouts. If the speed of sound in air is 340 m/s,
what is the time taken for the girl to hear the echo?
11. In a thunderstorm, both light and sound waves are produced at the same time.
a. Explain why you see light first before you hear the thunder.
b. What is the speed of:
i. Light?
ii. Sound in the air?
12. Describe an experiment that you would carry out to find the speed of sound in the air.
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13. Explain the difference between
a. natural and forced vibrations.
b. free and forced frequency.
14. a. Define resonance.
b. Explain the conditions for resonance to take place.
15. Describe an experiment that you would carry out to investigate resonance.
16. State two examples where resonance is:
a. Useful.
b. dangerous.
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Glossary
Absolute temperature: the minimum temperature that any substance can reach when it is cooled.
Acceleration: the rate of change of velocity per unit time.
Amplitude: the maximum displacement of an oscillating system from its resting
position.
Apparent weight: the weight of an object which is lower than the real weight of an object.
Audible sound range: the sound of the frequency which can be detected by human ear.
Circuit breaker: an automatic switch which, if the current rises over a specified value, the
electromagnet pulls the contacts apart, thereby breaking the circuit.
Component of a vector: projection of the vector on an axis.
Compressions: regions of high pressure and density along a longitudinal wave where
particles are squeezed.
Condensation: the process whereby a gas changes to liquid.
Constructive interference: a disturbance caused if two identical waves are in phase, both are moving
in the right direction
Coulomb: the electric charge which passes any point in a circuit in 1 second when a
steady current of 1 ampere is flowing.
Crests or peaks: points where a wave causes maximum positive displacement of the
medium.
Current electricity: the flow of electric charge.
Dependent variable: this is the variable which you measure in an investigation.
Destructive interference: a disturbance caused when two identical waves move in opposite
directions
Diffraction: the spreading out of waves when passing through a slit or a gap of an
obstacle.
Diffusion the movement of molecules (fluid molecules) from a region of high
concentration to a region of low concentration.
directions.
Displacement: how far and out of place an object is.
Displacement: the direction and distance from mean position. Displacement is measured
in metres (m) or centimeters (cm).
Distance: how much ground an object has covered during the motion.
Echo: the reflected or bounced off sound wave.
Effort: the force applied to a machine to move a load.
Electric circuit: a conducting path in which electrons flow or electric current takes place.
Electric current: the flow of electric charges (electrons) from the negative side of an
electric field to the positive side.
Electric shock: the passing of electric current through the body.
Electrical resistance: the opposition to the flow of electrons in a wire or a circuit.
Electromagnetic waves: the waves that do not need a medium for propagation.
Electromotive force (EMF): the maximum potential difference across a cell or battery when it not in a
305
circuit and not supplying current (when1=0 A).
Eureka can: the displacement can in Archimedes’ principle experiment.
Free fall: the falling of an object with uniform acceleration under the force of
gravity if air resistance is negligible.
Freezing: the process whereby a liquid is changed to a solid.
Frequency: the number of complete oscillations or cycles produced in a unit time.
Fuse rating: the maximum amount of current that a fuse can allow to pass through
before it melts.
Fuse: a component which is used to control the amount of current flowing in
the circuit.
Heat: the measure of the total internal energy contained in a body.
Inclined plane: a plane surface at an angle to the horizontal.
Independent variable: the variable that you are changing in an investigation or experiment.
Interference: a disturbance caused if two identical sets of waves travelling through the
same region of water resulting in either reinforcement or cancelling each
other.
Internal resistance (r): the resistance of a cell or battery to a current it causes.
is not in a circuit and not supplying current (when I = 0 A).
Kilowatt-hour: the electrical energy supplied in 1 hour to an appliance whose power is
1kw.
Kinetic energy: form of energy that a body possesses because of its motion.
Kinetic theory: a scientific explanation of the behaviour of the three states of matter.
Latent heat: the heat taken in or given out whenever a substance changes its states.
Lever: any rigid body which is pivoted about a point called the fulcrum.
Longitudinal wave: a wave in which the direction of the vibrating particles (oscillations) is
the same as the direction of a wave itself OR it is a wave in which the
displacements are parallel to the direction of a wave itself.
Machine: any device in which a force applied at one point can be used to overcome
a force at some other point.
matter.
Matter: anything which has mass and volume or occupies space.
Mechanical advantage: the ratio of the two forces, the load and the effort.
Mechanical waves: the waves that require a medium for propagation. They cannot pass
through a vacuum.
Melting: the process whereby a solid changes to a liquid.
Ohm: the electrical resistance of a conductor in which the current is 1 A when a
p.d. of 1 V is applied across it.
Oscillations or vibrations: complete upward or downward movements of an object about its fixed
position (rest position or equilibrium position) or complete to and fro
movements of an object about its fixed position (rest position or
equilibrium position).
Parallax error: the apparent change in the position of an object due to the apparent
change in the position of your eyes.
Parallel circuit: a circuit in which components are connected in branches.
Period: the time taken for one complete oscillation or cycle to be performed.
306
Potential difference: the difference in potential between two points, equal to the energy
change when a unit electric charge moves from one place to another in an
electric field.
Power: the rate of doing work, or it is the electrical energy transferred per unit
time, or it is the rate at which energy is produced.
Pressure: the force exerted per unit area.
Pulley: a grooved rim (rims) mounted in a framework called a block.
Rarefactions: regions of low pressure and density along a longitudinal wave where
particles are spaced.
Reflection: the bouncing off of the waves when an obstacle is placed in their path.
Refraction: the bending of a wave when it changes its speed or velocity.
Refractive index: the ability of a transparent material to bend waves.
Resistor: a device that causes resistance.
Resonance: a phenomenon takes place when a body is made to vibrate at its natural
frequency by vibrations received from another vibrating source of the
same frequency.
Resultant Vector: the final vector which is found when adding or subtracting vectors.
Scalar quantities: quantities that only give the magnitude (size or numerical value).
Series circuit: a circuit in which all the components are connected in one line.
Short circuit: the accidental touching of a live wire and a neutral wire.
Speed: the distance covered per unit time.
Standard notation/ scientific notation/ standard form: numbers written using powers of 10.
Temperature: The measure of the level of heat energy (measure of how hot or cold a
body is).
Tolerance: the extent to which the actual value of the resistance can vary.
Transverse wave: a wave in which direction of vibrations or oscillations is perpendicular
(at right angle) to the direction of propagation of the wave.
Troughs: points where a wave causes maximum negative displacement of the
medium.
Ultrasonic sound range: the sound of frequency that cannot be detected by human ear but other
animals e.g. dogs, bats and fish.
Upthrust: upward force which is exerted by a fluid on an object.
Variables: factors that would affect the results of the investigation.
Vector quantities: quantities that have both magnitude and direction.
Velocity ratio: the ratio of the distance moved by the effort to the distance moved by the
load in the same time.
Velocity: the distance covered in a stated direction (displacement) in a unit time.
Wave front: a line or section taken through an advancing wave which joins all points
which are in the same position in their oscillations.
Wave motion: the transmission of energy from one place to another through a material
or vacuum.
Wave phase: the orientation of wave pulses in space with respect to the origin of the
wave.
307
Wave: disturbance or oscillation that travels through a medium or vacuum,
accompanied by a transfer of energy.
Work done: force x distance moved by force in the direction of the force.
Work input: work put into the machine by the effort.
Work output: work done by the machine on the load.
Zero error: the error which occurs when the measuring instrument does not indicate
zero when it should.
308
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